Question

A girl having a mass of $35kg$ sits on a trolley of mass $5kg$. The trolley is given an initial velocity of $4\frac{m}{s}$ by applying a force. The trolley comes to rest after traversing a distance of $16m$.

$\left(a\right)$ How much work is done on the trolley?

$\left(b\right)$ How much work is done by the girl?

Answer 1

Step 1: Given data

The initial velocity of the trolley $\left(u\right)=4\frac{m}{s}$

The final velocity of the trolley $\left(v\right)=0$

Mass of the trolley $\left(m\right)=5kg$

Distance covered by the trolley before coming to rest $\left(s\right)=16m$

Step 2: Formula used

The third equation of motion $v^2=u^2+2as$

The formula for force $F=m\times a$

The formula for work done $W=F\times s$

Step 3: Calculating acceleration

From the third equation of motion

$$\begin{gathered} \Rightarrow v^2=u^2+2as \\ \Rightarrow 0=16+2a\left(16\right) \\ \Rightarrow a=-0.5\frac{m}{s^2}\left[Negativesignshowsretardation\right] \end{gathered}$$

Step 4: Calculating force.

Force of friction acting on the trolley,

$$\begin{gathered} \Rightarrow F=m\times a \\ \Rightarrow F=40\times -0.5 \\ \Rightarrow F=-20N\left[Negativesignshowsthatforceisactingintheoppositedirectionofmotion\right] \end{gathered}$$

Part $\left(a\right)$:

Work done on the trolley $=-$Work done by trolley

Work done on the trolley$=-F\times s$

$\Rightarrow W=20\times 16=320J$

Part $\left(b\right)$:

Since the girl is at rest with respect to the trolley,

So, the work done by the girl is $=0J$

Hence, the work done by the trolley and the girl is $320J$ and $0J$ respectively.