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?

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Solution

## Step 1: Given dataThe 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 usedThe third equation of motion ${v}^{2}={u}^{2}+2as$The formula for force $F=m×a$The formula for work done $W=F×s$Step 3: Calculating accelerationFrom the third equation of motion$⇒{v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}⇒0=16+2a\left(16\right)\phantom{\rule{0ex}{0ex}}⇒a=-0.5\frac{m}{{s}^{2}}\left[Negativesignshowsretardation\right]$Step 4: Calculating force.Force of friction acting on the trolley, $⇒F=m×a\phantom{\rule{0ex}{0ex}}⇒F=40×-0.5\phantom{\rule{0ex}{0ex}}⇒F=-20N\left[Negativesignshowsthatforceisactingintheoppositedirectionofmotion\right]$Part $\left(a\right)$:Work done on the trolley $=-$Work done by trolleyWork done on the trolley$=-F×s$$⇒W=20×16=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.

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