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InterviewSolution
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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 =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. 1600 JB. 160 JC. 16 JD. 1.6 J |
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Answer» Correct Answer - C (c ) Speed of the body, `v=(dx)/(dt)=(d)/(dt)((t^(3))/(3))=t^(2)` At`" "` t=0, v=0, At`" " t=0 s, v=4 ms^(-1)` From work energy theorem, W=change in kinetic energy `K_(f)-K_(i)` `=(1)/(2)m(v_(f)^(2)-v_(i)^(2))=(1)/(2)xx2xx(16-0)=16J` |
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