Under the action of a force, a 2kg body moves such that its position x as a function of time is given by x=t33, where x is in metre and t in seconds. The work done by the force in the first two seconds is :
A
1600J
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B
160J
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C
16J
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D
1.6J
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Solution
The correct option is A16J v=dxdt=t2 W=12mv2=12mt4 =12×2×(2)4=16J