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Question

Under the action of a force, a 2 kg body moves such that its position x axis as a function of time t is given by x=t23, where x is in metre and t in second. The work done by the force in first two seconds is

A
1600 J
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B
160 J
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C
16 J
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D
169 J
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Solution

The correct option is D 169 J
Here the position of the particle is given by the equation,
x=t23,
As we know that velocity is the time derivative of position.
v=dxdt=2t3
Applying the Work energy theorem
Final velocity vt=2=43 m/s and initial velocity is 0
Work done in first two seconds,
W=ΔK.E.=12(2)[(43)20]=169 J

Hence option D is the correct answer

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