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Question

The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t=x+3 where x is in metre and t in second. Calculate the work done by the force in the first 6s in Joule :

A
10
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B
0
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C
5
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D
2
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Solution

The correct option is B 0
Given equation can be rewritten as,
x=(t3)2
v=dxdt=2(t3)
vf=2(63)=6
vi=2(03)=6
So, from work-energy theorem
W=ΔKE=12m[v2fv2i]=12m[62(6)2]=0
i.e., work done by the force in the first 6s is zero.

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