Under the action of a force, a 2kg 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
1600J
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B
160J
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C
16J
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D
169J
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Solution
The correct option is D169J 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=43m/s and initial velocity is 0
Work done in first two seconds, W=ΔK.E.=12(2)[(43)2−0]=169J