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Q.

A force acts on a 2 kg object so that its position is given as a function of time as  x=3t2+5 , where x is in  meter and t is in second, then the work done by this force in first 5 seconds is  N×100 J then N = ________.

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answer is 9.

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Detailed Solution

V=dxdt=6 t

V(t=0)=0

V(t=5)=30 m/s

ΔkE=12×2×302=900J

N×100=900 ⇒N=9

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