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Question

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x=t33, where x is in meter and t in second. The work done by the force in the first two seconds is (Give an integer value)

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Solution

Given,

x=t33

Now, by differentiating the above equation, we get the velocity at ant time t as

v=dxdt=t2

At t=0, v=0 and

At t=2 s, v=4 m/s

From work energy theorem,

Work one by the force(W)=change in kinetic energy

W=K.EfK.Ei

W=12M(v2fv2i)

W=12×2×(4202)

W=16 J

Accepted answer : 16

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