A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

Answer:

Initial velocity of the trolley (u) = 4 m/s

Final velocity of the trolley (v) = 0

Mass of the trolley (m) = 5 kg

Distance covered by the trolley before coming to rest (s) = 16 m

From the equation

2as = v2 – u2

a = v2 – u2 /2s

a = (0 – 42)/2 × 16

a = 16/32

a = 0.5 m/s2

Frictional force acting on the trolley = m × a

F = 40 × (- 0.5)

F = – 20 N

(a) Work done on the trolley = – Work done by the trolley

= – Fs

= – (-20 N) x (16 m)

Work done on the trolley = 320 J

(b) Since the girl does not move w.r.t. the trolley (as she is sitting on it), work done by the girl = 0.

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