A girl having mass of $35 \mathrm{~kg}$ sits on a trolley of mass $5 \mathrm{~kg}$. The trolley is given an initial velocity of $4 \mathrm{~ms}^{-1}$ by applying a force. The trolley comes to rest after travelling a distance of $16 \mathrm{~m}$.
(a) How much work is done on the trolley?
(b) How much work is done by the girl?
Here, $u=4 \mathrm{~ms}^{-1}, v=0, \mathrm{~S}=16 \mathrm{~m}$
Using, $\quad v^{2}-u^{2}=2 a S$, we get
$a=\frac{v^{2}-u^{2}}{2 S}=\frac{0-16}{2 \times 16}=-0.5 \mathrm{~ms}^{-2}$
Total mass, $m=35+5=40 \mathrm{~kg}$
$\therefore$ Force, $\quad F=m a=40 \times(-0-5)=-20 \mathrm{~N}$
(a) Work done on trolley $=$ Force $x$ distance
$=20 \times 16=320 \mathrm{~J}$
(b) Work done by the girl $=0$
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