Work Energy Power

JEE Main level, numerical answer, medium

57 of 183

Question 57 of 183

A solid circular disc of mass 50kg50kg rolls along a horizontal floor so that its center of mass has a speed of 0.4ms−10.4m{s}^{-1}. The absolute value of work done on the disc to stop it is ______ JJ.

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Answer: 6

Using work-energy theorem, we get
W=∆KEW=∆KE
⇒W=0−(12mv2+12Iω2)\Rightarrow W=0-\left(\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2}\right)
⇒W=0−12mv2(1+K2R2)\Rightarrow W=0-\dfrac{1}{2}m{v}^{2}\left(1+\dfrac{{K}^{2}}{{R}^{2}}\right)
=−12×50×0.42(1+12)=−6J=-\dfrac{1}{2}\times 50\times 0.{4}^{2}\left(1+\dfrac{1}{2}\right)=-6J
Absolute work =+6J=+6J
Therefore, required value of W=6JW=6J

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