Question

Question 23
A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of $4m{s}^{-1}$ 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 1

(a)Given,

$u=4m{s}^{-1}$, $v=0$, $s=16m$

From the second equation of motion we have,
$a=\frac{v^2-u^2}{2s}=-\frac{16}{2\times 16}=-\frac{1}{2}m{s}^{-2}$

We know that,
$Force=ma=40\times (-\frac{1}{2})=-20N$
Work done on the trolley $W=F.s=20N\times 16m=320J$

(b) Work done by the girl = 0 J.

Work done by the girl is zero as she is stationary with respect to the cart i.e.; her displacement with respect to the cart is zero.