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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