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–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?
Mass of girl = 35 kg, mass of trolley = 5kg
Total mass = 35 + 5 = 40 kg
Initial velocity, u = 4 m/s,
final velocity, v = 0,
distance, s = 16 m
Using equation, v^2 = u^2 + 2as
0 = (4)^2 + 2a (16) => 32a = - 16 => a = - 0.5 m/s^2
Force exerted by trolley, F = ma
= 40 x 0.5 = 20 N
(a) Work done on trolley, W = F.S => 20 x 16 = 320 J
(b) Work done by girl, W = F.S => mass of girl x retardation x S
= 35 x 0.5 x 16 = 280 J.
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