A particle of mass `m` moves with a variable velocity v, which changes with distance covered x along a straight line as `v=ksqrtx`, where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is

Options

A. (a) `(mk^4)/(t^2)`
B. (b) `(mk^4t^2)/(4)`
C. (c) `(mk^4t^2)/(8)`
D. (d) `(mk^4t^2)/(16)`

Answer

C

Explanation

Given `v=ksqrt(x)` or `(dx)/(dt)=ksqrtx` or `x^(-1/2)dx=kdt`
Integrating both sides, we get
`(x^(1/2))/(1/2)=kt+C` assuming `x(0)=0`
Therefore, `C=0`
`2sqrtx=ktimpliesx=(k^2t^2)/(4)` or `v=(k^2t)/(2)`
Therefore, work done,
`DeltaW=Increase i n KE`
`=1/2mv^2-1/2m(0)^2=1/2m[(k^2t)/(2)]^2=1/8mk^4t^2`