Question

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

Answer 1

Step 1: (a) Given data

Mass of the girl, ${\mathrm{m}}_{\mathrm{g}}=35kg$

Mass of the trolley, ${\mathrm{m}}_t=5kg$

Total mass of the girl and the trolley,$\mathrm{m}=40kg$

Initial velocity,$u=4m{s}^{-1}$

Final velocity, $v=0m{s}^{-1}$

Distance travelled, $S=16m$

Step 2

The retardation of the trolley.

The negative acceleration or the retardation can be found by using the relation $v^2=u^2+2aS$, where $a$ is the acceleration.

$\begin{array}{rcl}a& =& \frac{v^2-u^2}{2S}\\ & =& \frac{{\left(0m/s\right)}^2-{\left(4m/s\right)}^2}{2\times 16m}\\ & =& \frac{-16}{2\times 16}\\ & =& -0.5m{s}^{-2}\end{array}$

Calculate the force $F$ by using the relation $F=ma$:

$\begin{array}{rcl}F& =& ma\\ & =& 40kg\times -0.5m{s}^{-2}\\ & =& -20N\end{array}$

Calculate the work done $\mathrm{W}$ by the trolley:

$\begin{array}{rcl}\mathrm{W}& =& F·S\\ & =& -20N\times 16m\\ & =& -320J\end{array}$

We can ignore the negative sign because the work done with respect to the initial force applied is being determined.

Step 2: (b) The work done by the girl is zero since she is sitting in the trolley, and hence there is no displacement with respect to the trolley.