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 `v= ax^(3//2)`
`m=0.5 kg , a = 5m^(-1//2) s^(-1)` work done (W) = ?
We know that,
Acceleration `a_0=(dv)/(dt) = v(dv)/(dx) =ax^(3//2) (d)/(dx) (ax^(3//2))`
`= ax^(3//2) xx a xx3/2 xx x^(1//2) = 3/2 a^2 x^2`
Now , Force `=ma_0 = m(3)/2 a^2 x^2`
Work done `=int_(x=0)^(x=2) Fdx=int_(0)^(2) 3/2 ma^2 x^2dx`
`=3/2 ma^2 xx(x^3 //3)_0^2`
`=1/2 ma^2 xx 8 =1/2 xx (0.5) xx (25) xx8 = 50 J`