Kinematics

JEE Main level, numerical answer, medium

125 of 308

Question 125 of 308

A particle of mass 10g10gmoves in a straight line with retardation 2x,2x, where xx is the displacement in SISI units. Its loss of kinetic energy for above displacement is (10x)−nJ.{\left(\dfrac{10}{x}\right)}^{-n}J. The value of nn will be ________.

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Answer: 2

Acceleration can be written as, a=vdvdxa=v\dfrac{dv}{dx}.
Given: a=−2xa=-2x(-ve sign is due to retardation)
Therefore,
vdvdx=−2x⇒vdv=(−2x)dxv\dfrac{dv}{dx}=-2x \Rightarrow vdv=\left(-2x\right)dx
Integrating both sides, we get
12(vf2−vi2)=−2(x22)=x2⇒∣12m(vf2−vi2)∣=mx2\dfrac{1}{2}\left({v}_{f}^{2}-{v}_{i}^{2}\right)=-2\left(\dfrac{{x}^{2}}{2}\right)={x}^{2} \Rightarrow \left|\dfrac{1}{2}m\left({v}_{f}^{2}-{v}_{i}^{2}\right)\right|=m{x}^{2}
⇒∣ΔKE∣=0.01x2=(10x)−2\Rightarrow |\Delta KE|=0.01{x}^{2}={\left(\dfrac{10}{x}\right)}^{-2}
Hence, n=2n=2.

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