Laws of Motion

JEE Main level, single correct, medium

16 of 210

Question 16 of 210

A body of mass 500g500g moves along xx-axis such that it's velocity varies with displacement xx according to the relation v=10xms−1v=10\sqrt{x}m{s}^{-1} the force acting on the body is:

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Answer: (B) 25N25N

It is given that
v=10xms−1v=10\sqrt{x}m{s}^{-1}
Force is defined by F=mdvdt...(i)F=m\dfrac{dv}{dt}...(i)
Multiplying and dividing equation (i) by dxdx
F=mdvdtdxdx⇒F=mdxdtdvdx=mvdvdx...(ii)F=m\dfrac{dv}{dt}\dfrac{dx}{dx} \Rightarrow F=m\dfrac{dx}{dt}\dfrac{dv}{dx}=mv\dfrac{dv}{dx}...(ii)
Finding the values of the terms in equation (ii)
dvdx=102x=5x\dfrac{dv}{dx}=\dfrac{10}{2\sqrt{x}}=\dfrac{5}{\sqrt{x}}
vdvdx=a=10x×5x=50ms−2v\dfrac{dv}{dx}=a=10\sqrt{x}\times \dfrac{5}{\sqrt{x}}=50m{s}^{-2}
Hence, the force is
F=12×50=25NF=\dfrac{1}{2}\times 50=25N.

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