A      box is sliding down an incline of angle 20A      box is s

A box is sliding down an incline of angle 20 A box is sliding down an incline of angle 20 Solution Part A) The component of the weight acting down the incline is mgsin(angle) Apply F = ma F = mgsin(angle) = ma (mass cancels) (9.8)(sin 20) = a a = 3.35 m/s^2 Part B) Now there is friction. Friction = uFn Fn = mgcos(angle) The net force = mgsin(angle) - umgcos(angle) = ma (mass cancels (9.8)(sin 20) - .2(9.8)(cos 20) = a a = 1.51 m/s^2 Part C) In this case mgsin(angle) = umgcos(angle) (m and g cancel) sin(20) = u(cos20) u = tan(angle) u = .364