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

Options

A. 25 J
B. 50 J
C. 100 J
D. 150J

Answer

B

Explanation

Given : `v = kx^(3//2)`
Acceleration, `a = (dv)/(dt) = (dv)/(dx)(dx)/(dt) = v (dv)/(dx)`
As `v^2 = k^2 x^3 :. 2v (dv)/(dx) = 3k^2 x^2 :. a = 3/2 k^2 x^2`
Force, `F = ma = 3/2 mk^2 [(x^3)/(3)]_(0)^(2) = 3/2 xx 0.5 xx 5^2 xx 8/3 = 50 J`