A body of mass 0.5 kg travels in a straight line with velocity `v= kx^(3//2)` where `k=5m^(-1//2)s^(-1)`. The work done by the net force during its displacement from x=0 to x=2 m is

Options

A. 1.5 J
B. 50 J
C. 10 J
D. 100 J

Answer

B

Explanation

Given :m=0.5 kg, `v=kx^(3//2)` where , k=`5 m^(-1//2) s^(-1)`
Acceleration , `a=(dv)/(dt)=(dv)/(dx)(dx)/(dt)=v(dv)/(dx) " " (because v=(dx)/(dt))`
As `v^2=k^2 x^3`
Differentiating both sides with respect to x, we get
`2v(dv)/(dx)=3k^2x^2`
`therefore` Acceleration, `a=3/2k^2x^2`
Force, F=Mass x Acceleration =`3/2mk^2x^2`
Work done, `W=int Fdx=underset(0)overset(2)int 3/2 mk^2x^2dx`
`W=3/2mk^2[x^3/3]_0^2=3/6xx0.5xx5^2xx[2^3-0]=50 J`