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Question

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

(a) How much work is done on the trolley?

(b) How much work is done by the girl?


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Solution

Step 1: Given data

The initial velocity of the trolley (u)=4ms

The final velocity of the trolley (v)=0

Mass of the trolley (m)=5kg

Distance covered by the trolley before coming to rest (s)=16m

Step 2: Formula used

The third equation of motion v2=u2+2as

The formula for force F=m×a

The formula for work done W=F×s

Step 3: Calculating acceleration

From the third equation of motion

v2=u2+2as0=16+2a(16)a=-0.5ms2Negativesignshowsretardation

Step 4: Calculating force.

Force of friction acting on the trolley,

F=m×aF=40×-0.5F=-20NNegativesignshowsthatforceisactingintheoppositedirectionofmotion

Part (a):

Work done on the trolley =-Work done by trolley

Work done on the trolley=-F×s

W=20×16=320J

Part (b):

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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