Question

A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4m//s by applying a force. The trolley comes to rest after traversig a distance of 16m. (a) How much work is done on the trolley ? (b) How much work is done by the girl ?

Answer 1

(a) Initial velocity of the trollely, u = 4m//s
Final velocity of the trolley `upsilon = 0` (as it comes to rest)
Mass of the trolley, m = 5kg
Distance covered by the trolley before coming to rest, s =16m
From `2 as = upsilon^2 - u^2, a = (upsilon^2 - u^2)/(2s) = (0 - (4)^2)/(2xx16) = - 0.5 m//s^2`
Force (frictional)acting on the trolley = ma =40 (-0.5) = - 20 N
Work done on the trolley = Fs = (20 N) (16m) = 320 J
(as to overcome friction force of - 20 N, we have to apply a force + 20 N)
(b) Since the girl does not move w.r.t the trolley (as she is sitting on it), work done by the girl = 0.