Q.
under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x=t33 , where x is in metre and t is in second. The work done by the force in the first two seconds is
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a
1600 Joule
b
160 Joule
c
16 Joule
d
1.6 Joule
answer is C.
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Detailed Solution
given positiion x=t33⇒Velocity=dxdt=t2velocity of the body v=t2Work done=change in KEWork=12mv2 ; substitite for velocity W=12mt22 ; substitute t=2;m=2 (given)W=12×2×24W=16J=work done in first 2 second
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