Work Energy Power

JEE Main level, single correct, medium

18 of 183

Question 18 of 183

A body of mass 0.5kg0.5kg travels on straight line path with velocity v=(3x2+4)ms−1v=\left(3{x}^{2}+4\right)m{s}^{-1}. The net work done by the force during its displacement from x=0x=0 to x=2mx=2m is

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Answer: (B) 60J60J

Given: v=3x2+4v=3{x}^{2}+4
Initial and final velocities are vi=3(0)2+4=4ms−1{v}_{i}=3{\left(0\right)}^{2}+4=4m{s}^{-1}
and vf=3(2)2+4=16ms−1{v}_{f}=3{\left(2\right)}^{2}+4=16m{s}^{-1}
From work-energy theorem,
⇒ΔKE=12m(vf2−vi2)\Rightarrow \Delta KE=\dfrac{1}{2}m\left({v}_{f}^{2}-{v}_{i}^{2}\right)
⇒ΔKE=12×12(162−42)\Rightarrow \Delta KE=\dfrac{1}{2}\times \dfrac{1}{2}\left({16}^{2}-{4}^{2}\right)
⇒ΔKE=W=60J\Rightarrow \Delta KE=W=60J

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