DC Pandey Mechanics Volume 1

540 — Mechanics - I 9. N A cos 30° = 500 N ...(i) 8.2INTRODUCTORY EXERCISE ...(ii) N A sin 30° = N B NA 1. (a) a = Net pushing force Ans. Total mass 30° NB = 120 − 50 1+ 4 + 2 30° w = 50 kg = 500 N = 10 m/s2 120 N 1 kg R On solving these two equations, we get N A = 1000 N (b) a 3 ∴ 120 − R = 1 × a = 10 and NB = 500 N R = 110 N 3 10. Net force in vertical direction = 0 (c) Fnet = ma = (2) (10) a = 20 N 30° T 30° 2. T4 = 4g and T1 = (1) g as a = 0 Ans. a ∴ T4 = 4 T1 (F = 20 N) a N 3. a = Net pulling force = F + 10 − 20 Total mass 1+ 2 w = 10 m/s2 3 ∴ T cos 30° = w or T = 2 w Ans. 3 Ans. 4. mAg = (2) (g) + (2)g sin 30° as a = 0 11. w = 100 N ...(i) ∴ mA = 3 kg Ans. ∑ (forces in horizontal direction) = 0 ∴ T + f cos 30° = N sin 30° 5. Since surface is smooth, acceleration of both is g sin θ = 10sin 30° = 5 m/s2,down the plane. N 30° T The component of mg down the plane (= mg sin θ) Cf provides this acceleration. So, normal reaction will 30° be zero. 6. N = mg is given N a 4 30° mg − mg = ma 4 w = 100 newton ∴ a = 3g mg 4 ∑(Forces in vertical direction) = 0 ∴ N cos 30° + f sin 30° = 100 ...(ii) 7. a = Net pulling force ...(iii) ∑ (moment of all forces aboutC) = 0 Ans. Total mass ∴ (T ) (R) = f (R) or T = f = (3 + 4 − 2 − 1) g = 4 m/s2 Ans. On solving these three equations, we get 3+ 4+ 2+1 T = f = 26.8N 4 kg 40 − T1 = 4a ∴ T1 = 24 N Ans. and N = 100 newton 3 kg T1 + 30 − T2 = 3a ∴ T2 = 42 N Ans. 1 kg T3 − 10 = (1) (a) ∴ T3 = 14 N Ans.

Chapter 8 Laws of Motion — 541 8. (a) a = Net pushing force and (x2 − x4) + (x3 − x4) = l2 (length of second string) Total mass = 100 − 40 = 3 m/s2 or x2 + x3 − 2x4 = l2 …(ii) 6 + 4 + 10 On double differentiating with respect to time, we (b) Net force on any block = ma get (c) N − 40 = 10 a = 30 a1 + a4 = 0 …(iii) and a2 + a3 − 2a4 = 0 …(iv) 10 kg 40 Newton But a4 = − a1 [From Eq. (iii)] N We have, a2 + a3 + 2a1 = 0 a This is the required constraint relation between a1, ∴ N = 70 newton Ans. a2 and a3. 9. (a) T1 = 10a, T2 − T1 = 20a, 60 − T2 = 30a 2. In above solution, we have found that 10 kg T1 a2 + a3 + 2a1 = 0 a Similarly, we can find T1 20 kg T2 v2 + v3 + 2v1 = 0 Taking, upward direction as positive we are given a v1 = v2 = 1 m/s ∴ v3 = − 3 m/s T2 30 kg 60 N i.e. velocity of block 3 is 3 m/s (downwards). a a = Net pulling force = 60 3. 2 kg 2T = 2 (a) ...(i) Total mass 10 + 20 + 30 1 kg 10 − T = 1 (2a) ...(ii) = 1 m/s2 On solving these two equations, we get T = 10 N On solving, we get T1 = 10 N, T2 = 30 N Ans. 3 (b) T1 = 0, T2 = 20a, 60 − T2 = 30a T a = Net pulling force = 60 Total mass 20 + 30 2 kg 2 T 1 kg 2a = 1.2 m/s2 a On solving, we get Ans. T2 = 24 N mg = 10 N 8.3INTRODUCTORY EXERCISE and 2a = 2T = 20 m/s2 3 1. Points 1, 2, 3 and 4 are movable. Let their 4. = acceleration of 1 kg block displacements from the fixed dotted line be x1, x2, x3 and x4 (ignoring the length over the pulley) T TM M x1 x4 x3 a mg ...(i) 1 4 ...(ii) T = Ma x2 3 Mg − T = Ma Ans. 2 On solving these two equations, we get We have, …(i) a= g x1 + x4 = l1 (length of first string) 2 and T = Mg 2

542 — Mechanics - I 5. In the figure shown, 8. T − Mg sin 30° = Ma ...(i) a = Net pulling force = 30 − 20 2Mg − 2T = 2M  2a ...(ii) Total mass 3+ 2 = 2 m/s2 Solving these equations, we get a= g Ans. 3 P1 P2 ® Stationary 2T 2T 2T T M P2 Mg sin 30° M a 2M a 30° 2 Ta 2 Mg T 2 kg a 3 kg 2 kg T − 20 = 2 × a = 4 8.4INTRODUCTORY EXERCISE ∴ Now, T = 24 N 1. (a) FA = − mAaB = (4$j) N Mg = 2T ∴ M = 2T = 48 (b) FB = − mBaA = (−4i$) N g 10 2. Constant velocity means acceleration of frame is = 4.8 kg Ans. zero. ...(i) 6. T = m1 (2a) = (0.3) (2a) = 0.6a 8.5INTRODUCTORY EXERCISE ...(ii) F − 2T = m2a 1. Figure (a) N = mg = 40 N ∴ 2T = 0.4 − 0.2a Ans. On solving these two equations, we get F µsN = 24 N a = 2 m/s2 Ans. Since, F < µsN 7 ...(i) ...(ii) Block will remain stationary and m1 T 2T m2 f = F = 20 N 2a a Figure (b) N = mg = 20 N and 2T = 12 N µSN = 12 N 35 and µKN = 8 N Since, F > µSN , block will slide and kinetic friction 7. a3 = a + ar = 6 (= 8 N) will act. a2 = ar − a = 4 a`= F − f = 20 − 8 = 6 m /s2 On solving these two equations, we get m2 a = 1 m/s2 2. If θ ≤ angle of repose, the block is stationary, a a = 0, Fnet = 0 and f = mg sin θ. If θ > angle of repose, the block will move, 1 f = µmg cos θ a a = mg sin θ − µmg cos θ ar 2 3 ar m = g sin θ − µg cos θ and Fnet = ma Further, µ = tan α (α = angle of repose = 45°) = tan 45° = 1

Exercises LEVEL 1 = m1 g  2m2  = m2 g  2m1   m1 + m2   m1 + m2  Assertion and Reason = m1 g  m2 + m2  = m2 g  m1 + m1  1. If net force is zero but net torque ≠ 0, then it can  m1 + m2   m1 + m2  rotate. Since m1 > m2, T < m1 g but T > m2 g Similarly, 2. If three forces are above a straight line (say AB ), we can prove that then there should be some force below the line AB also to make their resultant = 0. AB T < m2 g and T > m1g 3. a1 = g sin θ + µ g cos θ if m2 > m1. Therefore under all conditions T lies between m1g and m2g. a2 = g sin θ − µ g cos θ 8. A frame moving with constant velocity (a = 0) is v inertial. a1 9. If vector sum of concurrent forces is zero, then all mg sin q forces can be assumed a pair of two equal and q opposite forces acting at one point. mmg cos q F1 F1 = F2 mg sin q m mg cos q F2 P a2 Now, if we take moment about any general point P, v then moment of F1 is clockwise and moment of F2 q (of same magnitude) is anti-clockwise. Therefore net moment is zero. ∴ a1 = 1 + µ for θ = 45° a2 1 − µ 10. N = mg − F sinθ 4. For (−2i$) m/s2 component of acceleration, a F θ horizontal force in − ve x- direction is required which can be provided only by the right vertical F cosθ = µN = µ(mg − F sinθ) ∴ F = µmg = f (θ) wall of the box. cosθ + µ sinθ 5. By increasing F1 limiting value of friction will …(i) increase. But it is not necessary that actual value of friction acting on block will increase. 7. If m1 > m2 For F to be minimum, denominator should be maximum. a = Net pulling force = (m1 − m2) g or d (cosθ + µ sinθ) = 0 Total mass m1 + m2 dθ FBD of m1 From here we get θ = tan−1(µ) = angle of friction. m1 g −T = m1 a = m1 (m1 − m2) g With this value of θ we can also find that minimum m1 + m2 value of force required from Eq. (i). ∴ T = 2 m1 m2 g 11. Friction (or you can say net force on man from (m1 + m2) ground) is in the direction of motion.

544 — Mechanics - I Single Correct Option T 7. t = 2S ∝ 1 a 1. a = Net pulling force aa C Total mass ∴ t1 = a2 = (2 + 2 − 2) g = g mg t2 a1 2+ 2+ 2 3 Ans. 2= g sin θ =1 FBD of C 1 g sinθ − µK g cos θ 1 − µK mg − T = ma = mg 3 as, sin θ = cos θ at 45° ∴ T = 2 mg On solving the above equation, we get 3 µK = 3 = 2 (20) = 13.3 N 4 3 8. F1 = mg sin θ + µ mg cos θ 2. a = mg − F = g − F F2 = mg sin θ − µ mg cos θ mm Given that F1 = 2 F2 ∴(mg sin θ + µmg cosθ) = 2(mg sinθ − µmg cos θ) F F = air resistance a On solving, we get tan θ = 3µ 9. During acceleration, mg v = a1 t1 = F1 t1 m mA > mB ∴ aA > aB and A will reach earlier ∴ F1 = mv ...(i) t1 ...(ii) 3. Only two forces are acting, mg and net contact force During retardation, (resultant of friction and normal reaction) from the inclined plane. Since the body is at rest. Therefore 0 = v − a2 t2 = v − F2 t2 these two forces should be equal and opposite. m ∴ Net contact force = mg (upwards) ∴ F2 = mv t2 4. TA = 10 g = 100 N If t1 = t2 then F1 = F2 TB cos 30° = TA t1 < t2 then F1 > F2 and t1 > t2 then F1 < F2 ∴ TB = 200 N 3 10. Angle of repose θ = tan−1 (µ) TB sin 30° = TC = tan−1  13 = 30° ∴ TC = 100 N 3 5. a = mg − T = g − T R 30° mm T amin = g − Tmax a h m 30° 2 mg g mg So, particle may be placed maximum upto 30°, as 3 = g − = shown in figure m3  3  1  6. a = Net pulling force = 10 × 10 − 5 × 10 = 10 m/s2 h = R − R cos 30° = − R  2 Total mass 10 + 5 3 5 kg T − 5 × 10 = 5 × a = 50 11. Net pulling force F1 = 15g − 5g ∴ 3 F1 = 10g = 100 N T = 200 N Net stopping force F2 = 0.2 × 5 × 10 = 10 N 3 Since F1 > F2, therefore the system will move (anticlockwise) with an acceleration, This is also the reading of spring balance.

Chapter 8 Laws of Motion — 545 a = F1 − F2 = 90 Acceleration of particle towards each other 15 + 5 + 5 25 = component of T towards each other particle = 3.6 m/s2 m 15 kg block = T cos 60° = (F / 3) (1/2) = F 15 × 10 − T2 = 15 × a = 15 × 3.6 m m 2 3m ∴ T2 = 96 N 15. µ = tan φ 5 kg block N = mg − F sin θ T1 − 5 × 10 = 5 × a = 5 × 3.6 ∴ T1 = 68 N F ∴ T1 = 68 = 17 q T2 96 24 m 12. Acceleration of block with respect to lift F cos θ = µN = (tan φ) (mg − F sin θ) Pseudo force = (downward) =  sin φ  (mg − F sin θ) cos φ ar mg+ma On solving this equation, we get q F = mg sin φ cos (θ − φ) ar = m (g + a) sin θ = (g + a) sin θ 16. N = mg = 40 N m µN = 0.2 × 40 = 8 N S = 1 ar t2 At t = 2 s, F = 4 N 2 Since F < µN ∴ f =F =4N ∴ t = 2S = 2L 17. S = 1 at2 t = 2S or t ∝ 1 ar (g + a) sin θ aa 2 13. Since the block is resting (not moving) ∴ t1 = a2 t2 a1 ∴ f = mg sin θ ≠ µS mg cos θ ∴ t = g sin θ − µg cos θ or m = f 2t g sin θ or g sin θ = 10 10 × sin 30° = 2 kg Ans. 14. At point P, On solving this equation, we get 60° µ = 3 tanθ Ans. T 4 + 18. amr = acceleration of man relative to rope – 30° F = am − ar T 30° P ∴ am = amr + ar mg T = (a) + (a) = 2a Now, T − mg = m (am) = m (2a) ∴ T = m (g + 2a) Ans. F = 2T cos 30° Note Man slides down with a deceleration a relative ∴ T= F to the rope. So, amr = + a not −a. 3

546 — Mechanics - I 19. Steady rate means, net force = 0 Now at the mid point TT a T F f (50 +25) g T −µ  M  g = M 2 2 3T = 75 g = 750 ∴ T = 250 N Ans. a =  M   g2 or ...(i) 2 N2 = w ...(ii) 20. N 1 = µN 2 Σ MB = 0 Putting µ = 1 , we get Ans. w  23 = N 1 (4) ...(iii) 2 ∴ T = Mg N1 2 A 23. Angle of friction θ = tan−1 (µ) or θ = tan−1  1  = 30° 3 N2 F a B Suppose the body is dragged by a force F acting at O mN2 w On solving these equations, we get an angle α with horizontal. Then, µ=3 8 Ans. N = mg − F sinα and F cosα = µN = µ (mg − F sinα ) 21. In the figure, N 2 is always equal to w or 250 N ∴ F = µ mg ...(i) cosα + µ sinα A N1 F is minimum when denominator is maximum or d (cosα + µ sinα ) = 0 dα N2 or − sin α + µ cos α = 0 or tanα = µ = 1 B O mN2 3 w = 250 N ∴ α = 30° ,the angle of friction θ ∴ Maximum value of friction available at B is ∴ At α = 30°, force needed is minimum. µ N 2 or 75 N. Substituting the values in Eq. (i) we have, 22. a = F − f  1  (25) (g) 3 M Fmin = ( = Mg − µ mg = g 3 /2) + (1/ 3) (1/2) M2 = 12.5 g Ans. = 12.5 kg f m Mg T Note From this example we can draw a conclusion 2 a that force needed to move the block is minimum when pulled at an angle equal to angle of friction.

Chapter 8 Laws of Motion — 547 24. N = mg = 40 N 29. Maximum friction between two, µN = 32 N fmax = µ N = µ mg Applied force F = 30 N < µN upper block moves only due to friction. Therefore Therefore friction force, its maximum acceleration may be, f = F = 30 N ∴ Net contact force from ground on block amax = fmax = µg m = N 2 + f 2 = 50 N Ans. Relative motion between them will start when common acceleration becomes µg. 25. Maximum value of friction between B and ground, ∴ µ g = Net force = at Total mass 2m = µN = µ (mA + mB ) g = (0.5) (2 + 8) (10) = 50 N ∴ t = 2µ mg Ans. Since applied force F = 25 N is less than 50 N. a Therefore system will not move and force of friction between A and B is zero. 30. a1 = g sin 30° = g /2 26. mg sin θ = (10) (10) (3/5) = 60 N a2 = g sin 60° = 3 g/2 This 60 N > 30 N. Therefore friction force f will act in upward direction. ar = |a1 − a2 | FN Angle between a1 and a2 is 30° ∴ ar = a12 + a22 − 2 a1 a2 cos 30° = g 2 f Subjective Questions x 1. y . R F F = N 2 + f 2 = net force by plane on the block. 3 N 60° 60° 27. If force applied by man is F. Then in first figure 60° force transferred to the block is F, while in second figure force transferred to the block is 2F. 10 N 60° F FF ∑ Fx = 0 ∴ F − 3 cos 60° − 10 sin 60° = 0 FF Ans. ∴ F = 10.16 N Ans. ∑ Fy = 0 ∴ R + 3 sin 60° − 10 cos 60° = 0 ∴ R = 2.4 N (i) (ii) 2. Resolving the tension T1 along horizontal and 28. Total upward force = 4  mg  = 2 mg vertical directions. As the body is in equilibrium, 2 FF FF 60° Total downward force = (m + m) g = 2 mg T1 ...(i) ∴ Net force = 0 T2 F = mg 4 kg wt. 2 T1 sin 60° = 4 × 9.8 N

548 — Mechanics - I T1 cos 60° = T2 ...(ii) ∑ M0 = 0 T1 = 4 × 9.8 = 4 × 9.8 × 2 Ans. ∴ 40  2l  =T (l sin 60°) or T = 40 N sin 60° 3 Ans. 3 = 45.26 N T T2 = T1 cos 60° = 45.26 × 0.5 V 60° = 22.63 N OH 3. (a) At P Q ∴ mg = 40 N Ans. F1 45° 6. H = 20 N and V = 20 N T P F2 3 w Net hinge force = H 2 + V 2 = 40 N 3 F2 = T cos 45° = T 2 aA At P, ...(i) w = T cos 45° = T ...(ii) T 2 BT At Q, F1 = T ...(iii) a 2 mg From these three equations, we can see that For A T cos 45° = ma F1 = F2 = w = T= 60 = 30 2N Ans. or T = 2 ma ...(i) 2 2 Ans. For B mg − T cos 45° = ma 4. Various forces acting on the ball are as shown in ∴ mg − ma = ma or a = g figure. The three concurrent forces are in 2 equilibrium. Using Lami’s theorem, Substituting in Eq. (i), we get T = mg A B 2 30° 60° 7. a = Net pulling force = 8 Total mass 1+ 2+ 1 T1 90° T2 120° O 150° = 2 m/s2 T 8 − TB = m1a = (1) (2) B w = 10N ∴ TB = 6 N T1 = T2 = 10 T TA = m2 a = (2) (2) sin 150° sin 120° sin 90° or T1 = T2 = 10 A sin 30° sin 60° 1 ∴ =4N 8. (a) T1 − 2g = 2a T1 ∴ T1 = 10 sin 30° = 10 × 0.5 Ans. =5N Ans. and T2 = 10 sin 60° = 10 × 3 ...(i) 0.2 m/s2 0.1 kg 2 ...(ii) 1.9 kg =5 3N 2g 5. H = T cos 60° = T ∴ T1 = 2 (g + a) = 2 (9.8 + 0.2) 2 V + T sin 60° = 40 = 20 N

Chapter 8 Laws of Motion — 549 (b) T2 − 5g = 5a For vertical equilibrium, T2 N2 = w …(ii) Taking moments about B, we get for equilibrium, N 1(4 cos 30° ) − w(2 cos 60° ) = 0 …(iii) 2.9 kg Here, w = 250N 0.2 kg 5g Solving these three equations, we get 1.9 kg 0.2 m/s2 f = 72.17 N ∴ T2 = 5 (g + a) and N 2 = 250 N = 5 (9.8 + 0.2) ∴ µ = f = 72.17 = 50 N N 2 250 9. (a) a = Net pulling force = 0.288 Ans. Total mass = 200 − 16 × 9.8 11. Constant velocity means net acceleration = 0. 16 Therefore, net force should be zero. Only two forces = 2.7 m/s2 T and mg are acting on the bob. So they should be equal and opposite. (b) 200 − 50 − T1 = 5a Asked angle θ = 30° ∴ T1 = 200 − 5 × 2.7 − 50 T = mg = (1) (10) = 10 N = 136.5 N 200 N T (c) T2 − 9g = 9a a q T2 5 kg mg 50 N 30° T1 12. T sin θ − mg sin 30° = ma = mg a 2 kg 2 7 kg Ta 9g q ∴ T2 = 9 (g + a) 30° = 9 (9.8 + 2.7) 30° = 112.5 N mg 10. In figure, AB is a ladder N1 A ∴ T sin θ = mg T cos θ = mg cos 30° of weight w which acts C ...(i) at its centre of gravity or T cos θ = 3 mg ...(ii) ...(i) 2 Ans. G. Solving Eqs. (i) and (ii), we get Ans. ∠ ABC = 60° G θ = tan−1  23 ∴ ∠ BAC = 30° and T = 7 mg Let N 1 be the reaction N2 W 2 of the wall, and N 2 the reaction of the ground. 60° f =5 7N B Force of friction f between the ladder and the ground acts along BC. For horizontal equilibrium, f = N1

550 — Mechanics - I 13. a = Net pulling force 15. a = Net pulling force Total mass Total mass = 2g − (1) g 2+1 a m1 = g = 10 m/s2 m2 33 30° After 1 s, v = at = 10 m/s Ma 3 At this moment string slacks (T `= 0) a = Mg m1 + m2 + M 10 m 1 kg 2 kg u = 0 ...(i) 33 g m1 N cos 30° = m1g s1 g s2 N sin 30° = m1a ...(ii) ...(iii) 30° N String is again tight when, s1 = s2 ∴ 10 t − 1 gt2 = 1 gt2 (g = 10 m/s2) a 32 2 30° On solving we get, m1 t=1s Ans. From Eqs. (ii) and (iii), we get 3 a = g tan 30° = g 3 14. Since, acceleration of block w.r.t. wedge (an Substituting this value in Eq. (i) we get, accelerating or non-inertial frame of reference) is to M = 6.83 kg be find out. 16. (a) With respect to box (Non-inertial) N Ans. a = 5 m/s2 u = 10 m/s x (due to pseudo force) θ mg + FP = mg + ma x = x0 + uxt + 1 axt2 2 θ = x0 + 10t − 2.5t2 Ans. (g + a) sin vx = ux + axt = 10 − 5t Ans. θ a net= (b) vx = 0 at 2s = t0 (say) ∴ To return to the original position, FBD of ‘block’ w.r.t. ‘wedge’ is shown in figure. time taken = 2t0 = 4 s Ans. The acceleration would had been g sin θ (down the 17. (a) In car's frame (non-inertial) plane) if the lift were stationary or when only weight (i.e. mg) acts downwards. ax = − 5 m/s2 (due to pseudo force) Here, downward force is m(g + a). ux = 0 ∴ Acceleration of the block (of course w.r.t. az = 0 wedge) will be (g + a) sin θ down the plane. uz = 10 m/s

Chapter 8 Laws of Motion — 551 Now apply, v = u + at and s = s0 + ut + 1 at2 = (6) (10)  3  2   2 in x and z-directions. = 52 N (b) In ground frame (inertial) (a) Force needed to keep the block stationary is ax = 0, ux = 0, az = 0 and uz = 10 m/s F1 = F − µSN (upwards) 18. Relative to car (non-inertial) = 52 − 18 a1 = 5 m/s2 ux = 10 m/s = 34 N (upwards) Ans. a2 = µx = 3 m/s2 F 1 msN ax a1 is due to pseudo force F ax = − (5 + 3) = −8 m / s2 60° Block will stop when (b) If the block moves downwards with constant vx = 0 = ux + axt = 10 − 8t velocity (a = 0, Fnet = 0), then kinetic friction or at t = 1.25 s will act in upward direction. So, for t ≤ 1.25 s ∴ Force needed, F2 = F − µKN (upwards) = 52 − 12 x = x0 + uxt + 1 axt2 2 = 40 N (upwards) Ans. = x0 + 10t − 4t2 F 2 mKN vx = ux + axt = 10 − 8t After this µs g > 5 m/s2 as µs > 0.5 Therefore, now the block remains stationary with 52 N 60° respect to car. 19. N (c) Kinetic friction will act in downward direction 37° f q F3 − 52 − 12 = ma = (6) (4) 37° a ∴ F3 = 88 N (upwards) Ans. 37° F3 mg a N cos 37° + f sin 37° = mg ...(i) 52 N N sin 37° − f cos 37° = ma ...(ii) 60° On solving these two equations, we get m KN = 12 N f = 3.6 m Ans. 21. As shown in Fig. when force F is applied at the end = 9 mg 25 of the string, the tension in the lower part of the 20. N = mg cos 60° a = 1 m/s2 = (6) (10)  12 = 30 N F TT µSN = 18 N F µKN = 12 N Driving force F = mg sin 60° = force responsible to move the body downwards

552 — Mechanics - I string is also F. If T is the tension in string 25. X A + 2X B + XC connecting the pulley and the block, then, Fix line T = 2F But T = ma = (200)(1) = 200 N XB T XC ∴ 2F = 200 N B or F = 100 N XA A 22. N = F = 40 N f C F C = constant on double differentiating with respect to x time, we get, N aA + 2aB + aC = 0 Ans. mg = 20 N 26. P2 2T1 = 2T2 Net moment about C = 0 ∴ T1 = T2 ...(i) ∴ Anti-clockwise moment of 1 kg T2 = (1) (a) ...(ii) f = clockwise moment of N 2 kg T1 − 20 = 2  ar − a2 ...(iii) ∴ (20)  220 = (40) ⋅ x 3 kg 30 − T1 = 3  ar + 2a or x = 5 cm Ans. ...(iv) 23. Force diagram on both sides is always similar. 1 kg T2 a T2 Therefore motion of both sides is always similar. For example, if monkey accelerates upwards, then T2 a T > 20 g. But same T is on RHS also. P1 2 TT Monkey Bananas P2 2T2 ar T1 a 2 20 g 20 g 2 kg T1 a r Therefore, bananas also accelerate upwards. 3 kg 24. TA = 3T , TB = T On solving these equations, we get 2T T1 = T2 = 120 N 11 2T T TB a1 = a = 120 m/s2 11 A a2 = ar − a = − 50 m/s2 In such type of problems 2 11 a∝ 1 T or a2 = 50 m/s2 (downwards) 11 ∴ |aB | = 3 |aA| or aB = − 3aA, as aA and aB are in opposite a3 = ar + a = 70 m/s2 (downwards) Ans. directions. 2 11

Chapter 8 Laws of Motion — 553 27. 2T − 50 = 5a ...(i) (b) Substituting value of 't' in Eq. (ii), either on ...(ii) RHS or on LHS, common velocity = 6 m/s Ans. 40 − T = 4 (2a) On solving these equations, we get (c) s1 = u1t + 1 a1t2 2 2T T = (2) (1) + 1 (4) (1)2 = 4 m a 5 kg 4 kg 2 a 2 Ans. Ans. s2 = u2 t + 1 a2 t2 2 50 N 40 N = (8) (1) + 1 (−2) (1)2 = 7 m 2 a = 10 m/s2 or g 7 7 Ans. 30. N = 20 N = applied force Ans. and 2a = 20 m/s2 or 2g and µS N = 16 N 7 7 Driving force µKN = 12 N (downwards) 28. a = f = µg = 3 m/s2 F = mg m Since, F > µS N , block will slide downwards and kinetic friction of 12 N will act in upward direction a ∴ a = F − µKN m (downwards) f = mmg = 20 − 12 = 4 m/s2 (downwards) Ans. 2 (a) Relative motion will stop when velocity of block also becomes 6 m/s by the above 31. fmax = µ N = µ mg acceleration. = 0.8 × 2 × 10 = 16 N v = at ∴ t= v = 6=2s Block will not move till driving force F = 2t a3 Ans. becomes 16 N. This force becomes 16 N in 8 s (b) S = 1 at2 = 1 (3) (2)2 = 6 m Ans. For t ≤ 8 s a=0 22 For t ≥ 8 s 29. 2 kg block has relative motion towards right. a = F − fmax = 2t − 16 ⇒ a = t − 8 m2 Therefore, maximum friction will acts on it towards left. i.e. a - t graph is a straight line of slope +1 and intercept −8. f = µN = (0.4) (1) (10) = 4 N Corresponding a - t graph is shown below a1 = 4 = 4 m/s2 1 a (m/s2) a2 = − 4 = − 2 m/s2 2 1 kg 2 m/s 45° t (s) a1 4N 8 + – 4N 32. N = mg = 60 N 2 kg 8 m/s µS N = 36 N a2 µK N = 24 N Driving force F = 4t (a) Relative motion between them will stop when, Block will move when, v1 = v2 ...(i) t=9s ∴ u1 + a1t = u2 + a2t ...(ii) F = µS N or 4t = 36 ⇒ ∴ 2 + 4t = 8 − 2t For t ≤ 9 s or t = 1 s Ans. a=0

554 — Mechanics - I For t ≥ 9 s where, F= force on cube ∴ F = (8$i − 4$j) − w At 9 s block will start moving. Therefore kinetic = (8i$ − 4$j) − (−10$j) friction will act = (8i$ + 6$j) ∴ a = F − µK N m = 4t − 24 or |F| = (8)2 + (6)2 6 = 10 N Ans.  2 4 = 3 t − 3. During motion of block, a component of its ∴ a - t graph is a straight line acceleration comes in the direction of mg cos θ. At t = 9 s , a = 2 m/s2 Therefore, The corresponding a - t graph is as shown in figure. a (m/s2) mg cos q Motion relative to wedge 2 mg cos θ > N t (s) 4. Maximum friction available to m2 is 9 ( fmax ) = µ m2 g LEVEL 2 Therefore maximum acceleration which can be Single Correct Option provided to m2 by friction, (without the help of 1. Maximum value of friction between A and B normal reaction from m1) is ( f1)max = µN 1 = µmAg = 0.3 × 50 × 10 amax = fmax = µg = 150 N m2 Maximum value of friction between B and ground If a > µ g, normal reaction from m1 (on m2) is non ( f2)max = µN 2 = µ (mA + mB ) g zero. = (0.3) (120) (10) = 360 N 5. a = cot θ Force diagram is as shown below g T1 a q g q Ans. C ∴ a = g cot θ T1 B 2T2 A T2 6. x + y = constant mcg f2 f1 f1 v1 x A T2 = ( f1)max = 150 N y v0 T1 = 2T2 + ( f1)max + ( f2)max B v2 = 300 + 150 + 360 = 810 N T1 = mc g ∴ dx + dy = 0 dt dt ∴ mc = T1 = 810 = 81 kg Ans. g 10 2. a = dv = (8i$ − 4t$j) or  − ddxt  =  ddyt dt ∴ v1 − v0 = v2 Ans. At 1 s Fnet = ma = (1) (8$i − 4$j) = (8i$ − 4$j) or v1 − v2 = v0 =W + F

Chapter 8 Laws of Motion — 555 7. a1 = m2 g = 30 m/s2 or  − ddwt  + x  − ddxt  = dy ...(i) m1 + m2 7 z dt a2 = (m1 − m2) g − dw dx m1 + m2 dt dt = − = v2 = 10 m/s2 dy = v1 7 dt a3 = m2 g − m1g sin 30° and x = sinθ m1 + m2 z = 10 m/s2 Substituting these values in Eq. (i) we have 7 v2 (1 + sin θ) = v1 Ans. ∴ a1 > a2 = a3 Ans. 8. T − µ mg = ma 11. On the cylinder a If N = normal reaction between cylinder and T inclined plane m N sin θ = horizontal component of N ....(i) mmg = ma mmg N cos θ = vertical component of N = mg T m F ...(ii) a Dividing Eq. (i) by (ii) we get, ∴ T = µ mg + ma tan θ = a F − T − µ mg = ma g ∴ F − µ mg − ma − µ mg = ma ∴ a = g tan θ Ans. or a = F −µg Ans. 12. With respect to trolley means, assume trolley at rest 2m 9. a1 = sinθ and apply a pseudo force (= ma, towards left) on the a2 bob. q a2 a1 T q ma q ∴ a1 = a2 sin θ q anet mg 10. z = x2 + c2 x yz anet = mg sin θ − ma cos θ c m q = g sin θ − a cos θ Ans. w ...(i) 13. a = Net pulling force = (M − m) g Total mass (M + m) Ans. Since, M >> m Now, w+ y+ z=l ∴ M −m≈M +m=M or w + y + x2 + c2 = l Substituting in Eq. (i), we have a≈g ∴ dw + dy + x ⋅ dx = 0 FBD of m T − mg = ma = mg dt dt x2 + c2 dt ∴ T = 2 mg

556 — Mechanics - I 14. Let α = angle of repose = (10  3  − (10)  21 3)  2 For θ ≤ α Block is stationary and force of friction,  f = mg sin θ or f ∝ sin θ = 10 m/s2 i.e. it is sine graph For θ ≥ α Block slides downwards t = 2s = 2 × 1 = 1 s Ans. ar 10 5 ∴ f = µ mg cos θ or f ∝ cos θ a2 i.e. now it is cosine graph N The correct alternative is therefore (b). q 15. If they do not slip, then net force on system = F (Pseudo force) ∴ Acceleration of system a = F = ma 3m q T =F FBD of m F − µ mg = ma = F mg 3 18. ( f1)max = µ1m1g = 6 N a ( f2)max = µ 2m2g = 10 N At t = 2 s, F ′ = 4 N T=F Net pulling force, F1 = F`− F′ = 11 N mmg T = 5N 15N ∴ µ mg = 2F or µ = 2F Ans. f1 = 1N 3 3mg T = 5N 16. FBD of m w.r.t. chamber f2 = 10N Relative acceleration along the inclined plane Total maximum resisting force, ar = ma cos θ + mg sin θ F2 = ( f1)max + ( f2)max = 16 N m Since F1 < F2, system will not move and free body diagrams of the two block are as shown in figure. = (a cos θ + g sin θ) 19. a = Net pulling force t = 2s = 2L Ans. Total mass ar a cos θ + g sin θ = 2mg sin 30° = g N 2m + m 3 ma Tm (pseudo q a force) T = ma = mg /3 FBD of m 30° P q a2 mg 17. FBD of m w.r.t. wedge T Relative acceleration along the inclined plane T ar = ma cos θ − mg sin θ Resultant of tensions m R = − (T cos 30°) i$ − (T sin 30° + T ) $j = a cos θ − g sin θ

Chapter 8 Laws of Motion — 557 Putting T = mg /3 ∴ Maximum value of acceleration of coin which can be provided by the friction, R = − 3 mg i$ − mg $j 62 amax = fL = µS g ...(ii) m Since, pulley P is in equilibrium. Therefore, F+ R=0 Equating Eqs. (i) and (ii), we get where, F = force applied by clamp on pulley 4 = µS (10) ∴ F = − R = mg ( 3 i$ + 3 $j) Ans. ∴ µS = 0.4 Ans. 6 23. Just at the time of tipping, normal reaction at 1 will 20. ( f2 kg )max = µ2m2 g = 0.6 × 2 × 10 = 12 N become zero. ( f4 kg )max = µ4m4 g = 0.3 × 4 × 10 = 12 N Σ (moments of all forces about point 2) = 0 Net pulling force F = 16 N and N2 Net resistive force F′ = ( f2 kg )max + ( f4 kg )max = 24 N 12 Since, F < F′, system will not move and free body w1 w2 diagrams of two blocks are as shown below. 4m x TT F = 16N 2 kg 4 kg f2 f4 = 12N ∴ w1 (4) = w2 (x) Ans. or x = 4w1 = 4 (10 g) = 1 m 4 kg T + 12 = 16 ⇒ T = 4 N w2 80 g 2 2 kg f2 = T = 4 N Ans. 24. In vertical direction, net force = 0 21. FBD of rod w.r. t. trolley q N1 N1 a Pseudo force C N2 = ma q mg q mg ∴ N 1 cos θ = mg or N 1 = mg ...(i) cos θ N 1 = ma ...(i) N 2 = mg ...(ii) Under normal condition, normal reaction is, Net torque about point C = 0 Ans. N 2 = mg cos θ ...(ii) ∴ N1  l sin θ = N2  l cos θ ...(i) ∴ N 2 = cos2 θ Ans. 2 2 N1 or (ma) (sin θ) = (mg) (cos θ) 25. Σ (moments of all forces about point C ) = 0 ⇒ a = g cot θ N = mg cos q 22. v = 2t2 a = dv = 4 t C f = mg sin q ∴ dt At t = 1 s, a = 4 m/s2 mg sin q q O x mg cosq Limiting value of static friction, q = 45° fL = µS mg

558 — Mechanics - I ∴ f  2a = N (x) Net force in OP direction is zero ...(i) ∴ T + ma = mg or (mg sin θ)  2a = (mg cos θ) x 22 At 45° , sin θ = cos θ FBD of box w.r. t. ground ∴ x=a a 2 45° Hence, the normal reaction passes through O. T N = mg mg F= 3 C T cos 45° = ma or T = 2 ma 26. x Substituting in Eq. (i), we get f=F mg a= g mg 3 = 3 Ans. 29. 9 = tan 37° = 3 Σ (moments of all forces about point C ) = 0 ∴ Nx = F ⋅ a + f ⋅ a a4 ∴ a = 12 m/s2 22 or (mg) x = mg  2a + mg  2a Let N = normal reaction between rod and wedge. 3 3 Then N sin 37° will provide the necessary ma force or x = a to the wedge 3 Ans. a 27. x = l − l = l NB a 24 4 9 m/s2 37° NA x Cy w ∴ N sin 37° = ma = (10) (12) = 120 ∴ N = 120 = 120 y= l − l = l 263 sin 37° 0.6 Σ (moments of all forces about point C ) = 0 = 200 N Ans. ∴ NAx = NB y 30. N = ma N or N A = y = 4 a Ans. fL = µN = 0.5 ma fL NB x 3 Vertical acceleration, Ans. av 28. Let acceleration of box at this instant is 'a' (towards av = mg − fL = g − 0.5a mg m right). = 10 − 0.5 × 4 FBD of ball w.r.t. box O = 8 m/s2 pseudo T P Applying s = 1 at2 force 45° 2 = ma 45° anet mg or t = 2s in vertical direction, we have a t = 2 × 1 = 0.5 s 8

Chapter 8 Laws of Motion — 559 31. T − (nm + M ) g = (nm + M ) a 34. Maximum value of friction between 10 kg and 20 kg ∴ n (mg + ma) = T − Mg − Ma is a T 10 kg 100 N f1 f1 20 kg a 30 Kg (nm + M) a or n = T − M (g + a) ( f1)max = 0.5 × 10 × 10 = 50 N m (g + a) Maximum value of friction between 20 kg and 30 kg is nmax = Tmax − M (g + a) ( f2)max = (0.25) (10 + 20) (10) = 75 N m (g + a) Now let us first assume that 20 kg and 30 kg move as a single block with 10 kg block. So, let us first = 2 × 104 − 500 (10 + 2) calculate the requirement of f 1 for this 80 (10 + 2) 100 − f1 = 10 a = 14.58 Ans. f1 = 50 a But answer will be 14. On solving these two equations, we get 32. It implies that the given surface is the path of the f1 = 83.33 N given projectile Since, it is greater than ( f1)max, so there is slip y = x tan θ − gx2 θ between 10 kg and other two blocks and 50 N will 2u2 cos2 act here. (10) x2 Now let us check 50 N (20)2 cos2 = x tan 60° − 60° whether there is slip 20 kg a (2) between 20 kg and f2 30 kg or not. For this y = 3 x − 0.05 x2 ...(i) we will have to 30 kg f2 a Slope, dy = 3 − 0.1 x ...(ii) calculate dx requirement of f2 for no slip condition. At y = 5m 5 = 3 x − 0.05 x2 50 − f2 = 20 a and f2 = 30 a On solving these two equations, we get or 0.05 x2 − 3 x + 5 = 0 f2 = 30 N and a = 1 m/s2 x = 3 ± 3−1 = 3 ± 2 Since, f2 is less than ( f2)max, so there is no slip 0.1 0.1 between 20 kg and 30 kg and both move together From Eq. (ii) slope at these two points are, − 2 and with same acceleration of 1 m/s2. 2. 35. cos θ = 4R/5 = 0.8 33. Horizontal displacement of both is same (= l). R Horizontal force on A is complete T . But horizontal force on B is not complete T . It is component of T . So, horizontal acceleration of B will be less. T qR T 4R q R/5 5 θ = 37° ∴ v1 = tan θ = tan 37° = 3 v2 4 ∴ tB > tA

560 — Mechanics - I 2R ar = 2a ∴ a = ar = 2.5 m/s2 v1 2 q v2 q At point P 2T cos θ = F ∴ T= F 2 cos θ At mass m 3 3 Component of T along the other mass m is T sin θ 4 4 ∴ v1 = v2 = × 20 ∴ a=T sin θ = F  simnθ m  2 cos θ = 15 m/s Ans. 36. vAL = vA − vL + Lift ∴ F = 2 ma = 2 × 0.3 × 2.5 – tan θ (3/ 4 ) ∴ vA = vAL + vL = (−2) + 2 = 0 L =2N Ans. Let z = length of string at some 38. x + x + y2 + c2 = l = length of string instant. Then, Differentiating w.r.t. time, we get 2 m/s Fix x x Ö y2+ c2 x c B q y yA A Fixed B − dz = 2 m/s (Given) 2 dx = y c2  − ddyt dt Ans. dt y2 + Now, y = x − (z − x) = 2x − z or 2 vB = 1 vA ∴ cos θ dy dx  − ddzt dt = 2 dt + ∴ vA = 2 vB cos θ = (2) (10) (0.8) = 2 (2 m/s) + 2 m/s Ans. = 16 m/s = 6 m/s = vB 39. Maximum force of friction between c and ground is 37. Relative acceleration ( fc )max = (0.5) (60) (10) = 300 N Since it is pulling the blocks by the maximum force a (without moving). Therefore the applied force is m F = 300 N 3 cm q A 5 cm B F F = 300 N qP 4 cm q P T ( fAB )max = 0.4 × 60 × 120 × 10 = 240 N ( fBG )max = 0.3 × 120 × 10 = 360 N a m

Chapter 8 Laws of Motion — 561 Since ( fBG )max is greater than 300 N, blocks will not 43. Let a = maximum acceleration of A. move. Free body diagrams of block are as shown Under no slip condition acceleration of B is also a below. FBD of A w.r. t. ground A T=0 N y f=0 45° ax f=0 B T=0 45° F = 300 N f BG = 300 N 45° mN 40. Let aB = a$i aAB = aA − aB mg = (15 − a) i$ + 15$j Then, Σ Fy = 0 ...(i) ∴ N = mg + µ N ...(ii) Ans. 22 37° Σ Fx = ma aAB 37° ∴ N + µ N = ma 22 Solving these two equations, we get Since, aAB is along the plane as shown in figure. a = g  1 + µ  ∴ tan 37° = 3 = 15  1 − µ  4 15 − a 44. For M2 and M3 Solving this equation, we get a = − 5 2T or aB = (−5i$) 41. Acceleration, (towards left) TT a = 2F − F = F m + m 2m Horizontal forces on B gives the equation, a M2 a M3 2F − N sin 30° = m ⋅ a or 2F − N = m  2Fm a = M2g − M3g = 3M3g − M3g 2 M2 + M3 3M3 + M3 ∴ N = 3F Ans. =g 2 42. Distance AB = constant Now FBD of M2 gives the equation, M 2g B M 2g −T = M2⋅a = 2 75° ∴ T = M2g 60° 2 45° or 2T = M2g v Now taking moments of forces about support point A u M1g (l1) = (2T ) l2 = (M2g) (3l1) ∴ M1 = 3 ∴ Component of v along BA = component of u Ans. along BA M2 or v cos 60° = u⋅ cos 45° 45. Resultant of N and N (= 2 N )is equal to mg cos θ or v = 2 u Ans. ∴ 2 N = mg cos θ

562 — Mechanics - I ∴ N = mg cos θ 47. ( f1)max = between 1 kg and 2 kg 2 1 kg m = 0.2 NN 30 N 2 kg m = 0.5 1m Now kinetic friction will act from two sides = 0.2 × 1 × 10 = 2 N ∴ a = mg sin θ − 2µk N ( f2)max = between 2 kg and ground m a1 1 kg Substituting the value of N , we get 2N a = g (sin θ − 2 µk cos θ) Ans. 2 N a2 46. f1 → force of friction between 2 kg and 3 kg 1 kg 30 N ( f1)max = 0.5 × 3 × 10 = 15 N 15 N f2 → force of friction between 2 kg and 1 kg = 0.5 × 3 × 10 = 15 N ( f2)max = 0.3 × 5 × 10 = 15 N f3 → force of friction between 1 kg and ground a1 = 2 = 2 m/s2 1 ( f3)max = 0.1 × 6 × 10 = 6 N When F > 6 N system will start moving with a a2 = 30 − 15 − 2 2 common acceleration a= F −6 =  F − 1 m/s2 = 6.5 m/s2 3+ 2+1 6 ar = a2 − a1 a = 4.5 m/s2 f1 2 kg 1 kg t = 2Sr 6N ar − 6= (3) a = F − 3 = 2×1 = 2 s Ans. 2 4.5 3 f1 ∴ f1 =  6 + F − 3 More than One Correct Options 2 1. Maximum value of friction between two blocks  F 3 = 2 + F m2 T Since F is slightly greater than 6 N 2N ∴ f1 < 15 N or < ( f1)max 2 N m1 ∴ No slipping will occur here T a f2 fmax = 0.2 × 1 × 10 = 2N In critical case, 1 kg T =2N 6N F =T + 2=4 N ∴ System is in equilibrium if f ≤ 4 N f2 − 6= (1) (a) = F −1 Ans. 6 For F > 4 N ...(i) ∴ f2 = F + 5 F − (T + 2) = m2a = (1) (a) ...(ii) 6 T − 2 = m1 a = (1) (a) Again f2 < ( f2)max. So no slip will take place here also.

Chapter 8 Laws of Motion — 563 a 3. a = slope of v - t graph m2 T = −1 m/s2 F 2N ∴ Retardation = 1 m/s2 = µ mg = µ g 2N m m1 T or µ = 1 = 1 = 0.1 a g 10 On solving these two equations, we get If µ is half, then retardation a is also half. So using v = u − at T =F 2 or 0 = u − at or t = u or t ∝ 1 When, F = 6 N, T = 3 N Ans. aa 2. Resultant of mg and mg is 2 mg. we can see that t will be two times. T2 mg 4. Maximum force of friction between A and B b B ( f1)max = 0.3 × 60 × 10 = 180 N Maximum force of friction between B and ground ( f2)max = 0.3 × (60 + 40) g = 300 N mg A T = 125 N f1 Therefore T2 should be equal and opposite of this. f1 B T = 125 N f2 or T2 = 2 mg ...(i) Both are stationary Further, T2 cos β = mg ...(ii) f1 = T = 125 N ...(iii) f2 = T + f1 = 250 N and T2 sin β = mg or sin β = cos β ⇒ β = 45° 5. ax = mg sin θ = g sinθ m T1 cos α = mg + T2 cos β = mg + 2 mg  1  N xy 2 or T1 cos α = 2 mg ...(iv) T1 sin α = T2 sin β = 2 mg  1  mg sin q 2 mg cos q ∴ T1 sin α = mg ...(v) T1 It is also moving in y -direction a ∴ mg cos θ > N mg ay = mg cos θ − N m Bb mg ax a From Eqs. (iv) and (v), we get ay tan α = 1 and T1 = 5 mg 2 tan β = tan 45° = 1 and T2 = 2 mg ∴ tan β = 2 tan α and 2 T1 = 5 T2 Now, a = ax2 + a2y > g sin θ

564 — Mechanics - I 6. Maximum value of friction between A and B is 9. f1 → force of friction between 2 kg and 4 kg ( f1)max = 0.25 × 3 × 10 = 7.5 N f2 → force of friction between 4 kg and ground Maximum value of friction between B and C ( fS 1 )max = 0.4 × 2 × 10 = 8 N FK1 = 0.2 × 2 × 10 = 4 N ( f2)max = 0.25 × 7 × 10 = 17.5 N ( fS2 )max = 0.6 × 6 × 10 = 36 N and maximum value of friction between C and FK2 = 0.4 × 6 × 10 = 24 N ground, At t = 1 s, F = 2N < 36 N, therefore system remains ( f3)max = 0.25 × 15 × 10 = 37.5 N stationary and force of friction between 2 kg and F0 = force on A from rod 4 kg is zero. 7.5 N A F0 At t = 4 s, F = 8 N < 36 N. Therefore system is 17.5 N B 7.5 N again stationary and force of friction on 4 kg from F C ground is 8 N. T At t = 15 s, F = 30 N < 36 N and system is (f2)max = 17.5 N stationary. T 10. Net pulling force = 0 (f3)max = 37.5 N ⇒ a=0 If C is moving with constant velocity, then B will T1 = 1 × g = 10 N also move with constant velocity T3 = 2 × g = 20N T2 = 20 + T1 = 30 N For B, T = 17.5 + 7.5 = 25 N 11. fmax = 0.3 × 2 × 10 = 6 N For C, F = 17.5 + 25 + 37.5 = 80 N At t = 2 s, F = 2 N < fmax For F = 200 N ∴ f =F =2N Acceleration of B towards right At t = 8 s, F = 8 N > fmax = acceleration ofC towards left ∴ f =6N = a (say ) At t = 10 s, F = 10 N > fmax Then T − 7.5 − 17.5 = 4 a ...(i) ∴ f =6N a = F − f = 10 − 6 = 2 m/s2 200 − 17.5 − 37.5 − T = 8a ...(ii) m2 On solving these two equations, we get F = fmax = 6 N at 6 s For 6 s ≤ t ≤ 10 s a = 10 m/s2 a = F − f = t − 6 = 0.5t − 3 7. Since, µ1 > µ2 m2 ∴ ( f1)max > ( f2)max v 10 Further if both move, ∫ dv = ∫ adt = ∫ (0.5t − 3) dt a = T − µ mg m 06 µ of block is less. Therefore, its acceleration is more. v = 4 m/s 8. N cos θ = mg = 10 ...(i) After 10 s a = F − f = 10 − 6 = 2 m/s2 N sin θ = ma = 5 ...(ii) m2 On solving these two equations, we get = constant N ∴ v′ = v + at q = 4 + 2 (12 − 10) a = 8 m/s mg 12. Maximum force of friction between 2 kg and 4 kg N = 5 5 N and tan θ = 1 = 0.4 × 2 × 10 = 8 N 2

Chapter 8 Laws of Motion — 565 2 kg moves due to friction. Therefore its maximum 3. a = F + mg sin θ − µK mg cos θ acceleration may be m amax = 8 = 4 m/s2 (mg /6) + (mg /2) −  3 23  3  2 = mg   2 Slip will start when their combined acceleration  becomes 4 m/s2 m ∴ a = F or 4 = 2t or t = 12 s =g Ans. m6 3 At t = 3s 4. F′ = mg sin θ + µS mg cos θ a2 = a4 = F = 2t = 2 × 3 = mg + 4  3  m 6 6 mg  2 3 3  2 = 1 m/s2 = 7mg Ans. 6 Both a2 and a4 are towards right. Therefore pseudo forces F1 (on 2 kg from 4 kg) and F2 (on 4 kg from 5. F′′ = mg sin θ + µK mg cos θ 2 kg) are towards left = (mg /2) +  3 2   3  F1 = (2) (1) = 2N 3 mg  2 F2 = (4) (1) = 4 N  From here we can see that F1 and F2 do not make a pair of equal and opposite forces. = 5mg Ans. 6 13. See the hint of of Q.No-10 of Assertion and Reason 6. Acceleration a1 = 1350 × 9.8 − 1200 × 9.8 1200 type questions of Level-1. = 1.225 m/s2 vf = 0 Comprehension Based Questions Retardation, a2 = 1200 g − 1000 g h2 a2 1200 v 1. Let µK = µ, then µS = 2µ = 1.63 m/s2 h1 a1 According to first condition, F + mg sin θ = µS mg cos θ = 2 µ mg cos θ ...(i) h1 + h2 = 25 ...(i) According to second condition, v = 2a1 h1 or 2a2 h2 u=0 mg sin θ = F + µK mg cos θ ...(ii) or 2a1 h1 = 2a2 h2 = F + µ mg cos θ ∴ h1 = a2 Putting θ = 30°, we get h2 a1 ...(ii)  3  = 1.63 = 1.33 F + mg /2 = 2µ mg  1.225  2 or 3 µ mg = F + 0.5 mg ...(iii) Solving these equations, we get h1 = 14.3 m mg =F + µ mg  3  Ans.   2  2 7. v = 2a1h1 = 2 × 1.225 × 14.3 or 0.5 3 µ mg = 0.5 mg − F ...(iv) = 5.92 m/s Ans. Dividing Eq. (iii) and (iv), we get 8. tanθ = 8 ∴ θ = tan−1  185 = 28° F = mg 6 Ans. 15 2. Substituting value of F in Eq. (iii), we have ( fA)max = 0.2 × 170 × 10 × cos 28° = 300.2 N ≈ 300 N µ = 2 = µK 33 ( fB )max = 0.4 × 170 × 10 × cos 28° = 600.4 N ≈ 600 N ∴ µS = 2µ = 4 33 Now, (mA + mB ) g sin θ = (340) (10) sin 28° = 1596 N

566 — Mechanics - I Since this is greater than ( fA)max + ( fB )max, 2. (a) At θ = 0°, driving force F = 0 therefore blocks slides downward and maximum ∴ friction = 0 force of friction will act on both surfaces (b) At θ = 90° , N = 0 ∴ ftotal = ( fA)max + ( fB )max ∴ Maximum friction = µN = 0or friction = 0 (c) Angle of repose, = 900 N Ans. θr = tan−1 (µ) = 45° 9. a = (mA + mB ) g sin θ − ftotal F Since θ < θr, block is at rest and mA + mB f = mg sin θ = 2 × 10 sin 30° = 10 N (d) θ > θr. Therefore block will be moving = 1596 − 900 = 2.1 m/s2 340 f = µ mg cos θ = (1) (2) (10) cos 60° = 10 N a 4. (a) N − 10 = ma = 5 × 2 A (f A) max N 5 m/s2 g sin q mA F = force on connecting bar ms = 0.4 mk = 0.3 mA g sin θ − F − ( fA)max = mA a 10 N 2 kg ∴ F = mA g sin θ − ( fA)max − mA a = 170 × 10 × sin 28° − 300 − 170 × 2.1 F = 141 N Ans. w Match the Columns ∴ N = 20 N 1. F = 2t µsN = 8 N µkN = 6 N µs mg = 20 µs µk mg = 20 µk W = mg = 20 N (a) Motion starts at 4 s (b) When F = 15 N ∴ F = µs mg ⇒ (2) (4) = 20 µs w−F =5N (downwards) ∴ µs = 0.4 This is less than µs N (upwards) (b) At 4 s, when motion starts, ∴ f =5N a = F − µk mg m (c) F = w − µsN = 20 − 8 = 12 N ∴ 1 = 8 − 20 µk (d) F = w + µSN = 20 + 8 = 28 N 2 5. (a) Net pulling force F = net resisting frictional Solving we get, µk = 0.3 force at C = 10 N (c) At t = 0.1 s, when motion has not started, (b) fc = 0 (c) N c = (mB + mC ) g = 20 N (d) T = F = 10 N (everywhere) f = F = 2 × 0.1 = 0.2 N 7. Ground is smooth. So all blocks will move towards right 2 kg and 5 kg blocks due to friction. (d) At 8 s a = F − µk mg 2 kKgg m f1 = 2 × 8 − 0.3 × 20 f1 2 3 kg F = 5 m/s2 f2 ∴ a = 0.5 m /s2 f2 10 Ans. 5 Kkgg

Chapter 8 Laws of Motion — 567 8. µ mg cos θ = 15 N 2T 4T a a 4a T 2 B T Aa 15 N f mAg 10 N 30 N Similarly, if a be the acceleration of block A (downwards), then acceleration of block B towards T − 10 − 15 = 2a ...(i) right will be 4a. 30 − 2T = 3  a2 ...(ii) Equations of motion are Solving these two equations we get, For block A, mAg − 4T = mAa …(i) a = − 3.63 m/s2 or 50 − 4T = 5a For block B, T − f = 10(4a) So, if we take the other figure, or T − (0.1)(10)(10) = 40a or T − 10 = 40a …(ii) T 2T Solving Eqs. (i) and (ii), we get a 15 N a a = 2 m/s2 Ans. 2 33 10 N 30 N 3. (a) When the truck accelerates eastward force of friction on mass is eastwards. This figure is not feasible. Because for 'a'to be down frequired = mass × acceleration = (30 × 1.8) the plane, = 54 N 10 N > T + 15 Since it is less than µs mg (eastwards) which is not possible ∴ f = 54 N ∴ a=0 and free body diagrams are as shown below. (b) When the truck accelerates westwards, force of friction is westwards. T = 15 N 2T = 30 N frequired = mass × acceleration = 30 × 3.8 = 114 N a=0 Since it is greater than µsmg. Hence Ans. 30 N f = fk = µk mg = 60 N (westwards) a=0 4. Block B will fall vertically downwards and A along 10 N the plane. f=5N Writing the equations of motion. Subjective Questions For block B, …(i) mBg − N = mBaB …(ii) 1. It is just like a projectile motion with g to be or 60 − N = 6aB replaced by g sin 45°. (N + mAg) sin 30° = mAaA After 2 s, or (N + 150) = 30 aA v = vx2 + v2y Further aB = aA sin 30° 10 sin 45° − g 2 or aA = 2aB …(iii) 2 Solving these three equations, we get = × + (10 cos 45° )2 = 10 m/s Ans. (a) aA = 6.36 m /s2 Ans. (b) aBA = aA cos 30° = 5.5 m/s2 Ans. 2. Suppose T be the tension in the string attached to 5. Let acceleration of m be a1 (absolute) and that of M block B. Then tension in the string connected to block be a2 (absolute). A would be 4T .

568 — Mechanics - I Writing equations of motion. N N′ N′ N a1 f1 f1 f2 + F or f1 + f2 F mg cos a N sin a mg N + mg N + mg M a2 FBD of man FBD of plank For the plank not to move For m mg cos α − N = ma1 …(i) F − ( f2)max ≤ f1 ≤ F + ( f2)max …(ii) For M, N sin α = Ma2 or F − µ (M + m)g ≤ ma ≤ F + µ (M + m)g or a should lie between F − µ(M + m)g Note In the FBD only those forces which are along a1 and a2 have been shown. mm Constraint equation can be written as, and F + µ(M + m)g Ans. mm a1 = a2 sin α ...(iii) 8. Writing equations of motion Solving above three equations, we get 5T T acceleration of rod, a1 = mg cos α sin α Ans. + M  m sin α a1 a2  sin α  and acceleration of wedge mg FBD of m a2 = mg cos α Ans. Mg sin α + M FBD of M m sin α For M 5T − Mg = Ma1 …(i) 6. (a) N 2 and mg pass through G. N 1 has clockwise For m, mg − T = ma2 …(ii) moment about G, so the ladder has a tendency to From constraint equation, …(iii) slip by rotating clockwise and the force of friction a2 = 5a1 Solving these equations, we get ( f ) at B is then up the plane. (b) ΣMA = 0 acceleration of M , ∴ fl = mg  l sin 45° …(i) a1 =  5m −M  g 2  25m +M  ΣFV = 0 and of m, a2 = 5  5m − M  g  25m + M  ∴ mg = N 2 cos 45° + f sin 45° …(ii) From Eqs. (i) and (ii), 9. or a1 = s2 = n a2 s1 m N2 = 3 mg 2a1s1 = 2a2s2 22 A N1 g sin α =m and f = mg G or µg cos α − g sin α n 22 N2 f Solving it, we get mg B or µ min = f µ =  m + n tan α Ans. N2 m = 1 Ans. 45° 10. Limiting friction between A and B 3 fL = µN = 0.4 × 100 = 40 N 7. Here f1 = force of friction between man and plank (a) Both the blocks will have a tendency to move and f2 = force of friction between plank and surface. together with same acceleration (say a).

Chapter 8 Laws of Motion — 569 So, the force diagram is as shown. 12. Assuming that mass of truck >> mass of crate. a Retardation of truck a1 = (0.9) g = 9 m /s2 A F = 30 N f Retardation of crate a2 = (0.7) g = 7 m/s2 f B or relative acceleration of crate ar = 2m /s2 . Equations of motion are, 30 − f = 10 × a …(i) Truck will stop after time f = 25 × a …(ii) t1 = 15 = 1.67 s 9 Solving these two equations, we get a = 0.857 m/s2 and crate will strike the wall at and f = 21.42 N t2 = 2s = 2 × 3.2 = 1.78 s ar 2 As this force is less than fL, both the blocks will move together with same acceleration, As t2 > t1 , crate will come to rest after travelling a distance aA = aB = 0.857 m/s2 (b) 250 − f = 10a …(iii) 1 1  195 2 2 2 f = 25a …(iv) s = ar t12 = × 2.0 × Solving Eqs. (iii) and (iv), we get = 2.77 m Ans. f = 178.6 N 13. µkmg = 0.2 × 10 × 10 = 20 N a For t ≤ 0.2 s A F = 250 N Retardation a1 = F + µkmg f m f = 20 + 20 = 4 m/s2 B 10 As f > fL, slipping will take place between two At the end of 0.2 s, blocks and v = u − a1t f = fL = 40 N v = 1.2 − 4 × 0.2 = 0.4 m/s aA = 250 − 40 For t > 0.2 s 10 Retardation a2 = 10 + 20 = 3 m/s2 = 21.0 m /s2 10 Block will come to rest after time A F = 250 N t0 = v = 0.4 = 0.13 s 40 N a2 3 40 N ∴ Total time = 0.2 + 0.13 = 0.33 s Ans. B 14. Block will start moving at, F = µ mg aB = 40 = 1.6 m/s2 Ans. or 25t = (0.5) (10) (9.8) = 49 N 25 ∴ t = 1.96 s Velocity is maximum at the end of 4 second. 11. Normal reaction between A and B would be ∴ dv = 25t − 49 = 2.5t − 4.9 N = mg cos θ. Its horizontal component is N sin θ. dt 10 Therefore, tension in cord CD is equal to this ∫ ∫∴ vmax dv = 4 (2.5 t − 4.9) dt horizontal component. 0 1.96 Hence, T = N sin θ = (mg cosθ) (sin θ) Ans. ∴ vmax = 5.2 m /s Ans. = mg sin 2 θ 2

570 — Mechanics - I For 4 s < t < 7 s or a = M3 g M2 Net retardation a1 = 49 − 40 = 0.9 m /s2 10 ∴ F = (M1 + M2 + M3)a ∴ v = vmax − a1 t1 = 5.2 − 0.9 × 3 = 2.5 m /s = (M 1 + M2 + M3) M3 g Ans. For t > 7 s M2 Retardation a2 = 49 = 4.9 m /s2 18. Retardation a = µk g = 0.15 × 9.8 = 1.47 m/s2 10 Distance travelled before sliding stops is, ∴ t = v = 2.5 = 0.51 s s = v2 a2 4.9 2a = (5)2 ≈ 8.5 m ∴ Total time = (4 − 1.96) + (7 − 4) + (0.51) 2 × 1.47 Ans. = 5.55 s Ans. 15. Let B and C both move upwards (alongwith their 19. 2 N = mg cos θ pulleys) with speeds vB and vC then we can see that, ∴ N = mg cos θ √2 N N A will move downward with speed, 2vB + 2vC. So, 2 N with sign we can write, Ans. ∴ vB = vA − vc a = mg sin θ – 2µk N 2 m Substituting the values we have, vB = 0 Ans. = g sin θ – 2 µk g cos θ 16. FBD of A with respect to frame is shown in figure. = g (sin θ – 2µk cos θ) A is in equilibrium under three concurrent forces 20. v ⋅ dv = Net force = F − µk ρ (L − x) g shown in figure, so applying Lami's theorem N dx mass ρL 30° v L F − µk ρ (L − x) g dx mg v dv = Pseudo force A ∫ ∫∴ ρL = ma 60° 00 30° ∴ v2 = F − µk gL + µk gL 2 ρ 2 ma = mg ∴ v= 2F − µk gL Ans. sin (90 + 60) sin (90 + 30) ρ ∴ a = g cos 60° = 5.66 m/s2 21. (a) v = a1 t1 = 2.6 m/s cos 30° s1 = 1 a1t12 = 1 × 2 × (1.3)2 = 1.69 m 17. FBD of M2 and M3 in accelerated frame of reference 2 2 is shown in figure. s2 = (2.2 − 1.69) = 0.51 m Note Only the necessary forces have been shown. Now, s2 = v2 Mass M3 will neither rise nor fall if net pulling 2a2 force is zero. ∴ a2 = v2 = (2.6)2 = 6.63 m /s2 2s1 2 × 0.51 i.e. M2a = M3 g M2 and t2 = v = 0.4 s a2 Pseudo force = M2a (b) Acceleration of package will be 2m/s2 while M3 a retardation will be µk g or 2.5 m/s2 not M3g 6.63 m/s2. For the package,

Chapter 8 Laws of Motion — 571 v = a1t1 = 2.6 m /s ⇒ s1 = 1 a1 t12 = 1.69 m Relative retardation of upper block 2 ar = a1 + a2 or ar = 11 (µ 2 − µ1)g 1 1 10 s2 = vt2 − 2 a′2 t22 = 2.6 × 0.4 − 2 × 2.5 × (0.4 )2 Now, 0 = vm2 in − 2arl = 0.84 m ∴ vmin = 2arl = 22(µ2 − µ1)gl 10 ∴ Displacement of package w.r.t. belt Ans. Ans. = (0.84 − 0.51) m = 0.33 m Ans. (b) 0 = vmin − art Alternate Solution For last 0.4 s or t = vmin = 20l |ar | = 6.63 − 2.5 = 4.13m/s 2 ar 11(µ2 − µ1)g ∴ sr = 1 | ar |t 2 = 1 × 4.13 × (0.4 )2 24. vr = v12 + v22 2 2 Retardation a = µg = 0.33 m ∴ Time when slipping will stop is t = vr 22. Free body diagram of crate A w.r.t ground is shown a in figure. or t = v12 + v22 µg N …(i) aA sr = vr 2 = v12 + v22 2a 2 µg mAg = 100 N 30° xr = − sr cos θ = −  v12 + v22  v2   2µg      v12 + v22  Equation of motion is …(i) = −v2 v12 + v22 2 µg 100 − N = 10aA aA = a sin 30° = (2) 12 or aA = 1m/s2 yr = sr sin θ =  v12 + v22   v1       v12 + v22  Substituting in Eq. (i), we get  2 µg  N = 90 N. = v1 v12 + v22 2 µg 23. (a) Force of friction at different contacts are shown vr v1 = v1 in figure. m f1 10m f1 θ f2 –v2 = v2 Here, f1 = µ2mg In time t, belt will move a distance s = v2t and f2 = µ1 (11 mg) or v2 v12 + v22 in x-direction. Given that µ2 > 11µ1 µg ∴ f1 > f2 Retardation of upper block Hence, coordinate of particle, a1 = f1 = µ2g x = xr + s = v2 v12 + v22 m 2µg Acceleration of lower block v12 + v22 2 µg a2 = f1 − f2 = (µ2 − 11µ1)g and y = yr = v1 Ans. m 10

572 — Mechanics - I 25. FBD of m1 (showing only N a1 T T − mg = ma3 …(iv) T 2 the horizontal forces) …(i) From constraint equation, Equation of motion for m1 is a1 = a2 − a3 …(v) We have five unknowns. Solving the above five T − N = m1a1 equations, we get T T N T a3 NT a1 mg sin 30° =mg N a1 a3 a1 a2 2 FBD of 3m 2mg FBD of 2m FBD of m m2g a2 m3g FBD of m2 a1 = 3 g, a2 = 19 g and a3 = 13 g 17 34 34 Equations of motion for m2 are Acceleration of m = a3 = 13 g, N = m2a1 34 …(ii) and m2g − T = m2a2 …(iii) Acceleration of 2m = a12 + a22 = 397 g 34 Equation of motion for m3 are and acceleration of 3m = a1 = 3 g Ans. 17 m3g − T = m3a3 …(iv) Further from constraint equation we can find the 27. a = mAg relation, mA + M + m a1 = a2 + a3 …(v) For the equilibrium of B, We have five unknowns a1, a2, a3, T and N solving, mg = µN = µ(ma) = µmmAg we get mA + M + m a1 = (m2 + 2m1m3 g + m2m3 Ans. ∴ mA = (M + m)m m3) (m1 + m2 ) (µ − 1)m 26. Writing equations of motion, (M + m) µ − 1 T − N = 3ma1 …(i) mA = Ans. N = 2ma1 …(ii) …(iii) Note mA > 0 ∴ µ > 1 2mg − T = 2ma2

9. Work, Energy and Power INTRODUCTORY EXERCISE 9.1 −4 −4 1. W = F ⋅ S = F ⋅ (rf − ri ) 6. W = ∫ Fdx = ∫ (−2x) dx 22 = (6i$ − 2$j + k$ ) ⋅ [(2i$ + 3$j − 4k$ ) = [−x2 ]2−4 = − [16 − 4 ] = − 12 J Ans. Ans. − ($i + 4$j + 6k$ )] 2 24 4 x2 Fdx = 4 W∫ ∫7.= dx = − 2J = − 4 1 2 = − 4 1 − 1 = −1 J  x  4  2 4  2. (a) WF = FS cos 45° Ans. = (16) (2.2)  12 = 24.9 J 8. W = area under F -x graph (b) WN = N S cos 90° = 0 Ans. From X = − 4 to X = − 2 (c) Wmg = (mg) (S ) cos 90° = 0 Ans. F = − ve (d) Only three forces are acting. So, total work done (from graph) is summation of all above work done. S = + ve 3. WT = (T ) (x) cos 0° = Tx ∴ W1 = − 1 × 2 × 10 = − 10 J 2 WW = (W ) (x) cos 90° = 0 WN = (N ) (x) cos 90° = 0 From −2 to 4 WF = (F ) (x) cos 180° = − Fx F = + ve 4. mg − T = ma = mg T S = + ve 4 ∴ W2 = + 1 (6 + 2) (10) 2 ⇒ T = 3mg a 4 = + 40 J ∴ WT = (T ) (l) (cos 180° ) ∴ WT = W1 + W2 = 30 J Ans. = − 3 mgl mg 9. (a) From x = 10m to x = 5m 4 5. N = mg − F sin 45° = 18 − F S = − ve 2 F = + ve (from graph) Moving with constant speed means net force = 0 ∴ W1 = − Area = −5 × 3 F cos 45° = µN = 1 18 − F2  4 = − 15 J Ans. ∴ 4F = 18 − F (b) From x = 5m to x = 10 m 22 S = + ve ∴ F = 18 2 N 5 and F = + ve (a) WF = FS cos 45° ∴ W2 = + Area = 5 × 3  2  = 15 J Ans.  = 18 (2)  1  = 7.2 J (c) From x = 10m to x = 15 m 2  5 S = + ve F = + ve (b) W f = (µN ) (S ) cos 180° ∴ W3 = Area =  41 18 − F  (2) (−1) =3J Ans. 2 = − 7.2 J (d) From x = 0 to x = 15 m (c) Wmg = (mg) (S ) cos 90° = 0 S = + ve and F = + ve

574 — Mechanics - I ∴ W4 = + Area Ans. 7. (a) T = F, (b) WAll forces = 40 J = 1 × 3 × (12 + 6) = 27 J 8. v = ds = (4t − 2) 2 dt 10. (a) From x = 0 to x = 3.0 m Wall = ∆K = K f − Ki = K2 s − K0 s S = + ve F = + ve = 1 m (v 2 − vi2 ) ∴ W1 = + Area 2 f = + 4J = 1 × 2 [(4 × 2 − 2)2 − (4 × 0 − 2)2 ] (b) From x = 3 m to x = 4 m 2 F = 0 ⇒ W2 = 0 = 32 J Ans. (c) From x = 4 m to x = 7m 9. Wall = ∆K = K f − Ki = K f (as Ki = 0) S = + ve and F = − ve ∴ W3 = + Area = − 1 J ∴ Wmg + Wchain = K f (d) From x = 0 to x = 7 m or Wchain = Kf − Wmg = 1 mv2f − mgh W = W1 + W2 + W3 = 3 J 2 = 1 × 30 × (0.4)2 − 30 × 10 × 2 2 = − 597.6 J Ans. INTRODUCTORY EXERCISE 9.2 1. From work energy theorem, INTRODUCTORY EXERCISE 9.3 Wnet = Wmg + Wair = 1 m (v 2 − vi2 ) 1. ∆U = − W 2 f So, if work done by conservative force is positive ⇒ 0 + Wair = 1 × 0.1 [(6)2 − (10)2 ]= − 3.2 J then ∆U is negative or potential energy will 2 decrease. But there is no straight forward rule regarding the kinetic energy. 2. WAll = ∆ KE = K f − Ki 2. U A = − 60 J = 1 m (v 2 − vi2 ) = 1 × 2 × (0 − 202) 2 f 2 U B = − 20 J ∴ U B − U A = 40 J = − 400 J Ans. 4. vx = 0 = 0 : vx = b = α b ∴ WAll = ∆ KE = K f − Ki INTRODUCTORY EXERCISE 9.4 = 1 m (v2f − vi2 ) = 1 × m [(α b)2 − 0 ] 1. U = x3 − 4x + 6 2 2 3 = 1 m α 2b Ans. F = − dU = − x2 + 4 2 dx F = 0 at x = ± 2m 5. WF = F S cos 0° = 80 × 4 × 1 = 320 J S S Wmg = (mg) (S ) cos 180° = (50) (4) (−1) F x = –2 0 x=2 F = − 200 J F=0 F=0 K f = WAll = 120 J Ans. ∫6. K f − Ki = WF = Fdx For x > 2 m, F = − ve i.e. displacement is in positive direction and force is negative. Therefore 1 mv02 x x = 2 is stable equilibrium position. ∫0 − 2 = −Axdx For x < − 2 m, F = − ve 0 i.e. force and displacement are in negative directions. Therefore, x = − 2 m is unstable ∴ 1 mv02 = Ax2 or x = v0 m Ans. 2 2 A equilibrium position.

Chapter 9 Work, Energy and Power — 575 2. At A, x = 0 and F = 0 F For x > 0, F = + ve. i.e. force is in the direction of x=4m x>4m displacement. Hence A is unstable equilibrium F=0 F = +ve position. Same concept can be applied with E also. S At point C , F = 0. Therefore, equilibrium is unstable. For x > xC , F = − ve Displacement is positive and force is negative INTRODUCTORY EXERCISE 9.5 (in opposite direction of displacement). Therefore, C point is stable equilibrium point. 3. (a) At x = 0, F = 0. At P, attraction on −q or =W 1 mv2 1 m (at)2 t 2= 2 F1 > attraction F2 1. (a) Pav = t t x = -a Fnet F2 P F1 = 1 ma2 t = 1 × 1 × (4)2 (2) 22 x = +a +q x=0 –q +q = 16 W Ans. Ans. S (b) Pi = Fv = (ma) (at) = ma2t = (1) (4)2 (4) will get ∴ Net force is in the direction of displacement. Ans. So, equilibrium is unstable. = 64 W Ans. (b) Fnet is in opposite direction of S. Therefore 2. (i) W = Pt = 1 mv2 Ans. equilibrium is stable. 2 –q (ii) v = 2Pt FF m Fnet x = -a x = +a (iii) Integrating the velocity, we +q S +q displacement x=0 2P  t3/ 2  8P t3/ 2   ∴ S= = 4. U = minimum = − 20 J m  3/2 9m at x = 2 m Ans. 3. (a) K E = W = ∫ Pdt = ∫ 2t ⋅ dt = t2 ∴ x = 2 m is stable equilibrium position 1 mv2 = t2 5. F = (x − 4) 2 F = 0 at x = 4 m ∴ v= 2t m When x > 4 m, F = + ve When displaced from x = 4 m (towards positive (b) Pav =W = t2 =t t t direction) force also acts in the same direction. LEVEL 1 Exercises v = at = F t m Assertion and Reason ∴ But P = F⋅v = F2 t 1. F = constant a = F = constant m m ∴ P ≠ constant P∝t

576 — Mechanics - I 3. In the figure, work done by conservative force Single Correct Option (gravity force) is positive, potential energy is 3. (W f )A = + ve A FÞf A s decreasing. But kinetic energy may increase, B B f decrease or remain constant, depending on the value (W f )B = − ve of F. If there is no slip s F External force between A and B then f is static and total work done by static friction on system is zero.. 4. W = F ⋅ S = F ⋅ (rf − ri ) mg 55 4. In non-uniform circular motion (when speed ≠ 5. W = ∫ Fdx = ∫ (7 − 2x + 3x2) dx 00 constant) work done by all the forces is not zero. = 135 J Ans. 5. Slope of S -t graph is increasing. 6. Pi = F ⋅ v Therefore, speed of the particle is increasing. 7. P=W mgh + 1 mv2 =2 9. WAll = ∆K = K f − Ki tt = 1 m (v 2 − vi2 ) (800 × 10 × 10) + 1 × 800 × (20)2 2 f =2 vf = vt2 = Slope of S -t graph at t2 60 vi = vt1 = Slope of S -t graph at t1 = 4000 W Ans. These two slopes are not necessarily equal. 8. Ef = 80 % of Ei ⇒ mgh′ = (0.8) mgh 10. N Ans. or h′ = 0.8 h = 8m S2 q1 S2 q2 9. Ki + U i = K f + U f Sr N ∴ 0 + 6 = 1 × 1 × v2 + 2 q1 is acute S 2 q2 is obtuse 11. (W f1 )on A = − f1 S ∴ v = 2 2 m/s s (W f1 )on B = + f1 S f1 10. Maximum range is obtained at 45°. ∴ Net work done by f1 = 0 s E = 1 mu2 ...(i) B 2 Ans. f1 A At highest point, v = u cos 45° = u 2 1 1  u  2 2 2 2 ∴ E′ = mv2 = m =  1 mu2 = E 2 12. Decrease in mechanical energy in first case 22 = Ei − Ef = 1 mv2 − 0 11. W = Mgh + mg h =  M + m2  gh 2 2 = 1 mv2 h 2 2 Decrease in mechanical energy in second case h m 1 = Ei − Ef = 2 mv2 − mgh Further, µ does not depend on angle of inclination. M

Chapter 9 Work, Energy and Power — 577 12. (a) Velocity is decreasing. Therefore, acceleration = 50 − 50 = 100 N 33 (or net force) is opposite to the direction of motion. W = FS = 100 × 3 (b) and (c): some other forces (other than friction) 3 may also act which retard the motion. = 100 J 13. Let retarding force is F Ans. Then, Fx = 1 mv2 ...(i) 19. S = 0 (in vertical direction) 2 ...(ii) and F (x′ ) = 1 m (2v)2 ⇒ W = FS = 0 2 ∴ Pav =W =0 Solving these two equations, we get 20. k ∝ 1 t x′ = 4 x Ans. l 14. Ki + U i = K f + U f l of shorter part is less, therefore value of k is more W = 1 kx2 ∴ 1 mv02 + 0 = 1 mv2 + mgh 2 2 2 ∴ v = v02 − 2gh Ans. ∴ Wshorter part will be more. Ans. 15. a = F 21. Let θ1 is the angle of F with positive x-axis. m Fy 15 3 Fx 20 4 v = at =  mF  (2) ∴ tan θ1 = = = = m1 (say) P = F ⋅ v = 2F 2 Slope of given line, m2 = − α m 3 16. a = F W = 0 if F ⊥ S m or m1 m2 = − 1 ∴  43  − α3 = − 1 v = at = F t m  F 2 ∴ α =4 Ans.   P = F⋅v =  m t 22. Decrease in gravitational potential energy of block or P ∝ t = increase in spring potential energy i.e. P-t graph is a straight line passing through ∴ mg (xm sin θ) = 1 K xm2 origin. 2 17. K = 1 mv2 = 1 m (gt2) ∴ xm = 2 mg sin θ Ans. K 22 i.e. K ∝ t2 23. v = dS = (4 t) $i i.e. during downward motion. K -t graph is a dt parabola passing through origin with K increasing with time. Then in upward journey K will decrease P = F ⋅ v = 12 t2 with time. 22 18. Upthrust = (Volume immersed) (density of liquid) g ∴ W = ∫ Pdt = ∫ (12) t2dt 00 =  30500 (1000) (10) = 24 J Ans. 24. Decrease in potential energy = Work done against = 50 N friction 3 ∴ mg (h + d) = F ⋅ d weight = 50 N ∴ Applied force (upwards) = weight − upthrust Here F = average resistance ⇒ F = mg 1 + dh Ans.

578 — Mechanics - I 25. F = kx ⇒ x = F 30. Speed (and hence the kinetic energy) will increase k as long as mg sinθ > kx. Now, U = 1 kx2 31. (a) WN = NS cos 90° = 0 2  h  1  F  2 (b) WT = TS cos 0° = (mg sin θ)  sin θ 2 k = k or U ∝ 1 = mgh k (c) Wmg = (mg) (h) cos 180° = − mgh kB is double. Therefore, U B will be half. (d) WTotal = ∆K = K f − Ki = 0 As block is moved slowly or K f = Ki 26. 1 mv2 = 1 kX 2 2 2 m 32. Displacement of floor = 0 ∴ Xm = m ⋅v 33. W = area under F -x graph k From X = 1 to X = 3, force and displacement both = 0.5 × 1.5 Ans. are positive. Therefore work done is positive. 50 W1 = + Area = + 20 J = 0.15 m From X = 3 to X = 4, force is negative but displacement is positive. Therefore work done is 27. F = resistance is same again positive. 1 mv2 = n (F ⋅ d) 2 ∴ W2 = − Area = − 5J ⇒ n ∝ v2 From X = 4 to X = 5, force and displacement both are positive. Therefore work done is again positive. If v is doubled, n will become four times. ∴ W3 = + Area = + 5J 28. Ei = Ef WTotal = W1 + W2 + W3 = 20 J 0 = 1 × 10 × (0.15)2 − 0.1 × 10 × 0.15 2 34. v = dx = t2 + 1 × 0.1 × v2 2 dt ∴ v = 0.866 m/s Ans. W = 1 mv2 = 1 mt4 Ans. 22 29. From F = k x = 1 × 2 × (2)4 = 16 J 2 35. At highest point, d 1 mux2 = K 2 ∴ ux = 2K m 30° k = F = 100 R = 4H x 1.0 2uxuy = 4u2y = 100 N/m g 2g Decrease in gravitational potential energy ∴ uy = ux = 2K = increase in spring potential energy Now, m ⇒ 10 × 10 × (d + 2) sin 30° = 1 × 100 × (2)2 Ki = 1 mu2 = 1 m (ux2 + u2y ) 2 2 2 Solving, we get d = 2 m = 1 m  2K + 2K  2 m m ∴ Total distance covered before coming momentarily to rest = 2K = d + 2= 4 m Ans. Ans.

Chapter 9 Work, Energy and Power — 579 4 4. For increase in gravitational potential energy of a 36. ∆K = W = ∫ Pdt = ∫ (3t2 − 2t + 1) dt rod we see the centre of the rod. 2 Ans. = 46 J Ans. Ans. 37. K f − Ki = W = ∫ Fdx C⇒ 60° C 30 ∫∴ K f = Ki + ( − 0.1x) dx 20 W = change in potential energy = 1 × 10 × (10)2 −  x2  30 = mg l (1 − cos θ) 2 0.1  20 2  2  = 475 J Substituting the values, we have 38. KE = decrease in potential energy = mgh W = (0.5) (9.8)  12.0 (1 − cos 60° ) or K E ∝ m = 1.225 J Ans. or (KE)1 = 12 = 2 5. WT = (T ) (l) cos β β (KE)2 6 1 S=l T 39. W =U f −Ui =1K (X 2 − X 2 ) 2 f i N = 1 × 5 × 103 [(0.1)2 − (0.05)2 ] 90° + α 2 α = 18.75 J F W Subjective Questions WN = (N ) (l) cos 90° = 0 WW = (W ) (l) cos (90 + α ) = − W l sin α 1. K = p2 W f = (F ) (l) cos 180° = − F l 2m K′ = p′2 = (1.5p )2 6. Decrease in potential energy of chain 2m 2m = (2.25) p2 = 2.25 K = increase in kinetic energy 2m h= l ∴ % increase = K′ − K × 100 2 K C = 125 % Ans. ∴ mg l = 1 mv2 Ans. 22 1 or v = gl 2. p = 2Km or P ∝ K 2 7. T − mg = ma For small % changes, aT ∴ T = m (g + a) = 72 (9.8 + 0.98) mg % change in p = 1 (% change in K ) = 776.16 N 2 Ans. (a) WT = TS cos 0° = 1 (1 %) = 0.5 % = (776.16) (15) Ans. 2 = 11642 J 3. Total work done = − 1 K (2 x0 )2 = −2 K x02 (b) Wmg = (mg) (S ) cos 180° 2 = (72 × 9.8 × 15) (−1) ∴ Work done on one mass = − 10584 J = −2K x02 = − K x02 2

580 — Mechanics - I (c) K = WTotal = 11642 − 10584 11. FBD of particle w.r.t. sphere = 1058 J Ans. F = ma N mg (d) K = 1 mv2 Pseudo force 2 h = ma ∴ v= 2 × 1058 x 72 2K = θ a m vr R = 5.42 m/s Ans. 2 Ans. Kf = 1 mvr2 = WAll 2 8. (a) K = W = ∫ Fdx = ∫ (2.5 − x2) dx 0 = 2.33 J (b) Maximum kinetic energy of the block is at a or 1 mvr2 = WN + WF + Wmg point where force changes its direction. 2 or F = 0 at X = 2.5 = 0 + Fx + mgh = 1.58 m = (ma) (R sinθ) + mg [R (1 − cos θ)] 1.58 ∴ vr = 2gR (1 + sin θ − cos θ) Ans. ∫∴ Kmax = (2.5 − X 2) dx 12. From constraint relations, we can see that 0 = 2.635 J Ans. vA = 2 vB Therefore, vA = 2(0.3) = 0.6 m/s 9. W f = f S cos 45° as vB = 0.3 m/s (given) ApplyingWnc = ∆U + ∆K Here, f = mg sin θ, because uniform velocity means net acceleration = 0 or net force = 0 we get ∴ W f = (mg sin θ) (S ) (cos 45° ) −µ mAgS A = − mBgSB + 1 mAvA2 + 1 mB vB2 = (1) (10) (sin 45° ) (S ) (cos 45°) 2 2 If S = vt = 2 × 1 = 2 m Here, SA = 2SB = 2 m as SB = 1m (given) S ∴ − µ(4.0) (10) (2) = − (1) (10) (1) + 1 (4) (0.6)2 f 2 45° + 1 (1) (0.3)2 2 45° or −80µ = − 10 + 0.72 + 0.045 Wf = (1) (10)  1  (2)  1  or 80µ = 9.235 or µ = 0.115 Ans. 2 2 13. Let xmax = maximum extension of spring = 10 J Ans. Decrease in potential energy of A kxmax (say) mBg 10. vm1 = vm2 = v = Increase in elastic potential energy of Ans. spring dm1 = hm2 = 4m ∴ mA g x max = 1 kx 2 Using the equation, 2 max Ei − Ef =Work done against friction ∴ xmax = 2mA g k 1 (v2)  0 −  2 × (10 + 5) − 5 × 10 × 4  To just lift the block B, = 0.2 × 10 × 10 × 4 k x max = mB g Solving we get, ∴ 2mA g = mB g = mg v = 4m/s ∴ mA = m 2

Chapter 9 Work, Energy and Power — 581 14. (a) KA + U A = KB + U B (b) W f = f S cos 180° f ∴ 0 + 1 × 500 × (0.5 − 0.1)2 = (µmg cos θ) (S ) (−1) 180° 2 = 1 × 10 × v2 + 1 × 500 × (0.3 − 0.1)2 = −  21 (20) (cos 60° ) (2) S 22 On solving, we get = − 10 J Ans. v = 2.45 m/s Ans. 17. F = − dU = A Ans. dr r2 (b) CO = (BO )2 + (BC )2 18. (a) At x = 6 m, = (30)2 + (20)2 U = (6 − 4)2 − 16 = − 12 J ≈ 36 cm = 0.36 m K =8J Ans. ∴ E =U + K = −4 J Again applying the equation, (b) U min = − 16 J at x = 4 m Ans. KA + U A = KC + UC 0 + 1 × 500 × (0.5 − 0.1)2 ∴ Kmax = E − U min = − 4 + 16 2 = 12 J = 1 × 10 × v2 + 1 × 500 (0.36 − 0.1)2 22 On solving, we get Ans. (c) K = 0 v = 2.15 m/s ∴ U =E 15. Let X m is maximum extension of spring. Then or (x − 4)2 − 16 = − 4 decrease in potential energy of M = increase in or x = (4 ± 2 3) m Ans. elastic potential energy of spring (d) Fx = − dU = (8 − 2x) Ans. kxm dx µ mg cos 37° (e) Fx = 0 at x = 4 m Ans. mg sin 37° 19. Decrease in potential energy of 1 kg = increase in kinetic energy of both ∴ 1 kX 2 = MgX ⇒ 1 × 10 × 1 = 1 × (4 + 1) × v2 or 2 m 2 Now, or m ∴ v = 2 m/s Ans. ∴ Xm = 2 Mg 20. If A descends X then B will ascend 2x. Further if k 16. (a) speed of A at this instant is 2.5 m/s, then speed of B kX m = mg sin 37° + µmg cos 37° at this instant will be 5 m/s. Now, 2 Mg = (mg)  53 +  43 mg  45 Decrease in potential energy of A = increase in potential energy of B + increase in kinetic energy of M =3m both 5 Ans. ∴ (300) x = (50) (2x) + 1  390.80 (2.5)2 Ans. 2 Wmg = (mg) (S ) cos 30° 1  95.08 (5.0)2 = (20) (2) ( 3 /2) + 2 = 34.6 J Solving we get, x = 0.796 m Ans. 30° 21. If speed of sphere is v downwards then speed of S mg wedge at this instant will be v cot α in horizontal direction.

582 — Mechanics - I Now, v = 2a2 S2 = 1.5 Decrease in potential energy of sphere = Increase in ∴ S2 = 1.5 = 2 1.5 = 0.2 m kinetic energy of both 2a2 × 3.7 ∴ mgR = 1 mv2 + 1 m (v cot α )2 h = S2 sin 30° = 0.1 m 22 Now work done by friction = 1 mv2 cosec2 α = −[ initial mechanical energy] 2 = − mgh = − (0.05) (10) (0.1) ∴ v = 2gR sin α = speed of sphere and speed of wedge = v cot α = − 0.05 J Ans. = 2gR cos α Ans. 25. If A moves 1 m down the plane and its speed is v, 22. If block drops 12 mm, then spring will further then B will move 2m upwards and its speed will be 2v. stretch by 24 mm. Now, Ei = Ef dh ∴ 1 × 1050 × (0.075)2 = − 45 × 10 × 0.012 2 + 1 × 45 × v2 + 1 × 1050 × (0.099)2 θ 22 Solving we get, v = 0.37 m/s Ans. Using the equation, 23. If speed of block of 1.0 kg is 0.3 m/s then speed of Ei − Ef =Work done against friction 4.0 kg block at this instant would be 0.6 m/s. ∴ 0 −  1 × 30 × v2 + 1 × 5 × (2v)2 Applying,  2 2 Ei − Ef = work done against friction + 5 × 10 × 2 − 30 × 10 × 3 5  ∴ 0 −  1 × 1.0 × (0.3)2 + 1 × 4.0 × (0.6)2  2 2 = 0.2 × 30 × 10 × 4 × 1 − 1 × 10 × 1] = µk × 4 × 10 × (2) 5 Solving this equation we get, Solving this equation we get, µk = 0.12 Ans. v = 1.12 m/s Ans. 24. Retardation on horizontal surface, Note h = d sin θ = (1)  53 m=3 m 5 a1 = µg = 0.15 × 10 = 1.5 m/s2 Velocity just entering before horizontal surface, wA wB g g v = 2a1S1 mA = = 30 kg and mB = = 5 kg = 2 × 1.5 × 0.5 26. (a) Thermal energy = Work done against friction = 1.5 m/s = µK mgd = (0.25) (3.5) (9.8) (7.8) J a2 = 66.88 J S2 h Ans. 30° (b) Maximum kinetic energy Ans. = work done against friction Acceleration on inclined plane, = 66.88 J a2 = g sin θ − µg cos θ = 10 × 1 − 0.15 × 10 × 3 (c) 1 k xm2 = maximum kinetic energy 22 2 = 3.7 m/s 1 × 640 × xm2 = 66.88 2 xm = 0.457 m = 45.7 cm Ans.

Chapter 9 Work, Energy and Power — 583 LEVEL 2 = mgR 1 − π2 Single Correct Option Now, Ui + Ki =U f + Kf ∴ 1. WAll = 1 mv2 mgR 1 − 2  = 0+ 1 mv2 2 π 2 ∴ WF + Wmg + WN = 1 mv2 1 − 2  2 π or v= 2gR Ans.  1 5 1 1 ∴ (5 × 5) + 2 × 10 × + 0= 2 × 2 × v2 5. T = 2mg ∴ v = 14.14 m/s Ans. As soon as string is cut T (on A) suddenly becomes 2. P = F v = constant zero. Therefore a force of 2mg acting on upward ∴ F=P direction on A suddenly becomes zero. v So net force on it will become 2mg downwards. ∴ mv  ddvs = P ∴ a1 = 2mg = 2g (downwards) v m s m 2v Spring force does not become instantly zero. So ∫ ∫∴ ds = v2dv acceleration of B will not change abruptly. P or a2 = 0 0v 6. Let X m is maximum elongation of spring. Then, Solving we get, s = 7mv3 Ans. 3P increase in potential energy of spring = decrease in potential energy of C. 3. F = kx 1 ∴ k = F = 100 ∴ 2 KX 2 = M1 gX m x1 m = 100 N/m or KX m = maximum spring force Ei = Ef = 2M1g = µ min Mg ∴ 1 × 10 × v2 = 1 × 100 × (2)2 − (10) (10) (2 sin 30° ) ∴ µ min = 2M 1 Ans. 22 M Solving we get, 7. Ti = mg ...(i) v = 20 m/s Ans. 2kx = 2mg ∴ kx = mg 4. dm =  m dθ =  2πm dθ One kx force (acting in upward direction) is    π / 2 suddenly removed. So, net downward force on system will be kx or mg. Therefore, net downward h = R (1 − cos θ) acceleration of system, dθ a = mg = g θ 2m 2 Free body diagram of lower block gives the equation, mg − Tf = ma = mg 2 dm h ∴ Tf = mg ..(ii) 2 From these two equations, we get dU i = (dm) gh = 2mgR (1 − cos θ) dθ ∆T = mg Ans. π 2 ∫∴ Ui = π/2 = 2mgR  π − 1  8. Fnet = mg sin θ − µmg cos θ ...(i) π 2 dU i = mg sin θ − 0.3 xmg cos θ 0

584 — Mechanics - I At maximum speed Fnet = 0. Because after this net We can see that initial velocity is in the direction of force will become negative and speed will decrease. PO. So the particle will cross the X -axis at origin. From Eq. (i), Fnet = 0 at Ki + Ui = Kf + U f x = tan θ = 3/4 = 2.5 m 0.3 0.3 Ans. ∴ 0 + (3 × 6 + 4 × 8) = K f + (3 × 0 + 4 × 0) 9. At C, potential energy is minimum. So, it is stable or K f = 50 J Ans. equilibrium position. 14. F = 0 at x = x2. When displaced from x2 in negative Further, direction, force is positive i.e. in the opposite F = − dU = − (Slope of U -r graph) direction of displacement. Similarly, when dr displaced in positive direction, force is negative. Negative force means attraction and positive force 15. Fnet = mg sin θ − µmg cos θ means repulsion. 10. P = F ⋅ v = mg sin θ − µ0xg cos θ = (−mg $j)⋅[ux$i + (uy − gt) $j ] a = Fnet = g sin θ − µ0xg cos θ = (−mguy ) + mg2t m i.e. P versus t graph is a straight line with negative ∴ v⋅ dv = g sin θ − µ0 xg cos θ intercept and positive slope. dx 11. Ei − Ef = Work done against friction 0 xm ∴1 kx2 − 1 K  x2 2 = µmg  x + x2 or ∫ vdv = ∫ (g sin θ − µ0xg cos θ) dx 2 2 00 Solving this equation we get, ∴ x = 4µmg Ans. xm = 2 tan θ Ans. k µ0 12. P = F ⋅ v = Fv cos θ = TV cos θ Ans. 16. ∑ (Moments about C ) = 0 13. F = −  ∂φ i$ + ∂φ $j ∴ (k1x) AC = (k2x) BC  ∂x ∂y ∴ AC = k2  ...(i) BC k1 = (−3i$ − 4$j) Y AC + BC = l ...(ii) Ans. P Solving these two equation we get, 53° Ans. AC =  k2  l  k1 + k  8m  2  53° 17. Work done by friction = Ef − Ei O 6m X = 1 × 1 × (2)2 − 1 × 10 × 1 2 -3^i =−8J 53° 18. F = − dU = 12a − 6b dx x 13 x7 O -4^j At equilibrium, F = 0 or x =  2ba 1/6 Since, particle was initially at rest. So, it will move At this value of x, we can see that d 2U is positive. in the direction of force. dx2 So, potential energy is minimum or equilibrium is stable.

Chapter 9 Work, Energy and Power — 585 19. ∑ (Moment about O ) = 0 kx = mg or, x = mg k ∴ (kx) l = mg  2l  or x = mg 2k and this does not depend on h. U = 1 kx2 = (mg)2 Ans. 25. dm =  mπ  dθ 2 8k 20. Ei = Ef dm h = R sin θ h h=0 0= m2gh − m1gh + 1 (m1 + m2) v2 dθ 2 θ ∴ v= 2gh  m1 − m2  Ans.  m dθ  m1 +  π  m2 ∴ dU = (dm) gh = gr sin θ 21. F = − dU ∫U i = π = 2mgr π dx dU or dU = − Fdx = (ax − bx2) dx 0 AssumingU = 0 at x = 0, and integrating the above Now, Ki + Ui = Kf + U f equation we get, ∴ 0 + 2mgr = 1 mv2 − mg  π2r U = ax2 − bx3 π 2 23 ∴ v= 2gr  2 + π2  Ans. U = 0 at x = 0 and x = 3a π 2b 26. Maximum speed is at equilibrium where For x > 3a , b x3 > a x2 and U will become 2b 3 2 F = kx ⇒ x = F k negative. So, option (c) is the most appropriate answer. Now, F ⋅ x = 1 mv2 + 1 kx2 22 22. W = FS  F  1 1  F  2 k 2 2 k F and S are same. Therefore, or F = mv2 + k WA = 1 Solving we get, WB 1 F From work energy theorem, v= mk = vmax Ans. KA = WA = 1 27. Increase in potential energy per unit time = decrease KB WB 1 in kinetic energy of both 1 1 or 2 mAvA2 = 2 mB vB2 = − d  1 m1v12 + 1 m2v22 dt 2 2 ∴ vA = mB = 2 vB mA 1 = v1  −m1 dv1  + v2  −m2 dv2  dt dt 23. Work done by conservative force = v1 (−m1 a1) + v2 (−m2 a2) = − ∆U =Ui −Uf or dU = v1 (−F1) + v2 (−F2) ...(i) = [k (1 + 1)] − [k (2 + 3)] dt = − 3k Ans. Here, −F1 = − F2 = kx = 200 × 0.1 = 20N Substituting in Eq. (i) we have, 24. After falling on plank downward force on block is dU = (4) (20) + (6) (20) mg and upward force is kx. Kinetic energy will dt increase when mg > kx and it will decrease when kx > mg. Therefore it is maximum when, = 200 J/s Ans.

586 — Mechanics - I 28. There is slip, so maximum friction will act on A in = (3i$ + 4$j) m/s Ans. ...(i) the direction of motion (or towards right) | v| = (3)2 + (4)2 = 5 m/s S 2. F = − dU = 5 − 200x A dx f At origin, x = 0 ∴ F = 5N f = µ mg = 0.2 × 45 × 10 = 90 N Ans. S = 40 − 10 − 10 = 20 cm = 0.2 m a = F = 5 = 50 m/s2 ∴ W = f S = 18J m 0.1 29. Constant velocity means net force = 0 Mean position is at F = 0 or at, x = 5 = 0.025 m Using Lami's theorem in the figure, 200 N a = F = 5 − 200x = (50 − 2000x) f 37° m 0.1 53° 53° At 0.05 m from the origin, S = vt = 20 m x = + 0.05 m 37° or x = − 0.05 m mg = 100 N Substituting in Eq. (i), we have, We have, |a | = 150 m/s2 N = 100 or = 50 m/s2 sin (180° − 53°) sin 90° At 0.05 m from the mean position means, ∴ N = 100 sin 53° = 80 N x = 0.075 Now, WN = NS cos 53° or x = − 0.025 m = (80) (20) (0.6) = 960 J Ans. Substituting in Eq. (i) we have, Ans. | a | = 100 m/s2 30. In both cases sudden changes in force by cutting the 3. Spring force is always towards mean position. If spring would be kx. displacement is also towards mean position, F and S will be of same sign and work done will be positive. ∴ a = ∆F = kx (in both case) mm 4. Work done by conservative force = − ∆U In one case it is downwards and in other case it is Work done by all the forces = ∆K Work done by forces other than conservative forces upwards. = ∆E More than One Correct Options 5. At equilibrium 1. F = −  ∂U $i + ∂U $j = (−7$i − 24 $j) N m Equilibrium position  ∂X ∂y a = F =  − 7 i$ − 24 $j m/s m 5 5 |a| =  75 2 +  254 2 5 m/s2 Ans. 3m = Since, a = constant, we can apply, kδ0 = mg v = u + at or δ0 = mg k = (8.6i$ + 23.2$j) +  − 7 i$ − 24 $j 5 5 (4 ) where, δ0 = compression

Chapter 9 Work, Energy and Power — 587 (b) δTotal =δ + δ0 = 3 mg Solving these two equations, we get k a = 0 and T = 2mg Fmax = kδmax = k  3mk g = 3mg (downward) T a m T 2m a kx = 2 mg Nmax 2 mg 3m 8. Work done by conservative forces 3 mg Fmax = U i − U f = − 20 + 10 = − 10 J Work done by all the forces, ∴ N max = 3mg + Fmax = 6mg Ans. = K f − Ki = 20 − 10 = 10 J (d) If δ > 4mg, then upper block will move a k 9. (a) Work done by gravity in motion 1 is zero distance x > 4 mg − δ0 or x > 3mg from natural (θ = 90° ) and in motion 2 is negative (θ = 180° ). k k (b) In both cases angle between N and S is acute. length. (c) and (d) : Depending on the value of acceleration Hence in this case, extension in motion 1, friction may act up the plane or down the plane. Therefore angle between friction and x > 3mg displacement may be obtuse or acute. So, work done k by friction may be negative or positive. or F = kx > 3mg (upwards on lower block) Comprehension Based Questions So lower block will bounce up. 1. U = E − K = 25 − K 6. (a) Decrease in potential energy of B = increase in Since, K ≥0 ∴ U ≤ 25 J spring potential energy ∴ 2mg xm = 1 k xm2 2. U = E − K = − 40 − K 2 Since K ≥0 ∴ xm = 4 mg Ans. ∴ U ≤ − 40 J k (b) Ei = Ef Match the Columns 1. S = rf − ri = (+ 2 $i ) − (+ 4 $i ) = − 2$i 1 1  2mg  2 2 2 k Now apply W = F ⋅ S 0= (m + 2m) v2 + × k × 2. Friction force = 0, as tension will serve that purpose −(2mg )  2mg  k NS ∴ v = 2g m T 3k (upwards) mg (c) a = kxm − 2mg ...(i) 2m Angle between N and S or between T and S is acute = k  4mk g − 2mg = g ∴ WT and WN are positive. Angle between S and 2m mg is 180°. (d) T − 2mg = ma Therefore, Wmg is negative. 2mg − T = 2ma ...(ii)

588 — Mechanics - I 3. (a) Net force is towards the mean position x = 0 S = 1 at2 = 1 × 10 × (0.3)2 = 0.15 m 2 23 where, F = 0when displaced from this position. Therefore, equilibrium is stable. TT S Net force 1 kg 2 kg +q +q S x = 0 +Q S 10 N 20 N (b) Net force is away from the mean position. (a) Wmg on 2 kg block = (20) (S ) cos 0° Ans. Therefore, equilibrium is unstable. = (20) (0.15) = 3J Ans. Ans. Net force (b) Wmg on 1 kg block = (10) (S ) cos 180° Ans. S = (10) (0.15) (−1) Ans. = −1.5J +q x=0 +q (c) WT on 2 kg block = (T ) (S ) cos 180° (c) Same logic can be applied as was applied in =  430 (0.15) (−1) part (b). = −2 J (d) Net force is neither towards x = 0 nor away from x = 0. Therefore, equilibrium is none of the three. F2 F1 (d) WT on 1 kg block = (T ) (S ) cos 0° =  430 (0.15) (1) x=y Q q q =+2J x=0 6. (a) f 4. (a) From A to B speed (or kinetic energy) will be S increasing. Therefore net potential energy W f = f S cos 180° = negative should decrease. S=0 (b) From A to B, a part of decrease in gravitational potential energy goes in increasing the kinetic (b) energy and rest goes in increasing the potential energy of spring. f (c) and (d) : From B to C kinetic and gravitational wf = 0 potential energy are decreasing and spring potential energy is increasing. (c) and (d) : No solution is required. ∴ (decrease in kinetic energy) + (decrease in Subjective Questions m1 F gravitational potential energy) = increase in spring potential energy. 1. 5. Common acceleration of both blocks m2 a = Net pulling force T µ Total mass a (F − µm1g )xm = 1 kxm2 = 20 − 10 = 10 m/s2 2 1+ 2 3 or kxm = 2(F − µ m1g) 1 kg T − 10 = ma = 1 × 10 10 N Second block will shift if kxm ≥ µm2g 3 ∴ 2(F − µm1g) > µm2g ∴ T = 40 N or F >  m1 + m22 µg 3

Chapter 9 Work, Power and Energy — 589 2. x A h=0 dθ θ h θ B Final Initial Initial PE, Total mass being pulled = m (x + h) l θ = π/2 ∴ Acceleration a = Net pulling force ∫U i = θ = 0° (r dθ) (ρ) (g) (r cos θ) Total mass being pulled = (ρg r2) [sin θ ] π / 2 = ρgr2 = gh 0 x+h Final PE, Uf =  πr × ρ (g)  − πr/2  =− π 2 r2ρg ∴ v  − ddvx = gh 2 2 8 x+h ∆U = r2ρ g 1 + π 2 v0 dx  8 v ⋅ dv = − gh ∫ ∫or 0 (l − h) x + h ∆U = KE v2 1 + π 2 1  π2r ∴ 2 = gh [ln(x + h)]l − h  8 2 or r2ρg = (ρ) v2 0 ⇒ v= 4 rg  1 + π  or v2 = gh ln  hl  π 8 2 π 4 ∴ v = 2ghln(l / h) Ans. 2 π or v= rg  +  Ans. 5. (a) From energy conservation principle, 3. For t ≤ 0.2 s Work done against friction = decrease in elastic PE F = 800 N and v =  02.03 t or f (x0 + a1 ) = 1 k (x02 − a12 ) 2 ∴ P = Fv = (53.3 t) kW Ans. or x0 − a1 = 2f …(i) k For t > 0.2 s From Eq. (i), we see that decrease of amplitude F = 800 −  800.10 (t − 0.2) and v = 20 t (x0 − a1 ) is 2f , which is constant and same for each 0.3 k ∴ P = Fv = (160 t − 533 t2) kW cycle of oscillation. 0.2 0.3 (160 t − 533 t2) dt (53.3 t) dt + ∫ ∫W = 0 0.2 = 1.69 kJ Ans. 4. At the instant shown in figure, net pulling force x0 a1 = m gh l