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
A
1600 J
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B
160 J
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C
16 J
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D
1.6 J
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Solution
The correct option is C 16 J v=dxdt=t2
At t = 0, v = 0,
At t = 2 s, v = 4 m/s
From work energy theorem,
W = change in kinetic energy Kf−Ki =12m(v2f−v2i) =12×2×(16−0) =16J