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?

Given: Mass of the girl $m_{girl}=35\ kg$

Mass of trolley $m_{trolley}=5\ kg$

So, effective mass $m=m_{girl}+m_{trolley}=35\ kg+5\ kg=40\ kg$

Initial velocity $u=4\ ms^{-1}$

Distance travelled $s=16\ m$

Final velocity $v=0$

On using, third equation o motion, $v^2=u^2+2as$

Or $2as=v^2-u^2$

Or $2\times a\times 16=0-4^2$

Or $32a=-16$

Or $a=\frac{-16}{32}$

Or $a=-\frac{1}{2}\ ms^{-2}$       [$-ve$ sign indicates retardation]

Force applied, $F=ma=40\times(-\frac{1}{2})=-20\ N$

$(a)$. Work done on the trolley$=-$work done by the trolley

$=-(F.s)$

$=-(-20\ N\times16\ m)$

$=320\ Joule$

$(b)$. The girl is sitting on the trolley, so there is no displacement in her position. so, work done by the girl is zero.

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Updated on: 10-Oct-2022

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