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 travelling a distance of 16m. (a) How much work is done on the trolley ? (b) How much work is done by the girl ?

Explanation

Given, u=4m//s, v=0 and s=16m
From the third equation of motion, [:'" for retardation, the acceleration is negative" i.e., a=-a]
`v^(2)=u^(2)-2as`
`(0)^(2)=(4)^(2)-2axx16`
`0=16-32a`
`a=(16)/(32)=0.5ms^(-1)`
u= inital velocity, v = final velocity, a = acceleration and s = displacement
(a) Total mass = 35 +5 = 40kg
Work is done on the trolley W = F.d = ma s [:' F=ma]
`=40xx0.5xx16=320J`
(b) Mass of girl m 35kg
Work done by the girl `W=F.d=mas=35xx0.5xx16=280J`