Question

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

Answer 1

in each case, the object traveling has initial kinetic energy, and work is done by each on friction that equals this KE.

Answer 2

Mass of girl = 35 kg
Mass of trolley = 5 kg
Total mass, m = (35 + 5) = 40 kg
u = 4 m/s
v = 0
s = 16 m
By equation,
v 2 = u 2 + 2as
0 = (4)2 + 2a (16)
32a = –16
a = –0.5 m/s2
Force exerted on trolley,
F = ma
= 40 × 0.5 = 20 N

Work done on trolley W = FS = 20 N x 16 m
= 320 J
Work done by the girl, W = FS
= mass of girl × retardation × S
= 35 × 0.5 × 16
= 280 J

Answer 3

work done by girl is zero,because she is sitting on the trolley and force is perpendicular to displacement.

but in book it is given 280J.

How is it?