A body of mass 0.5 kg travels in a straight line with velocity `v=a x(3/2)` where `a=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?

Explanation

We are given:
Here m = 0.5 kg
`v = ax^(3//2)`
`a = 5 ms^(-2)`
Initial velocity at `x = 0,v_i = a xx 0 = 0`.
Final velocity at `x = 2, v_f = a (2)^(3//2) = 5(2)^(3//2)`.
According to work-energy theorem
Work done = change in K. E.
`= 1/2 m(v_f^2 - v_i^2) = 1/2 xx 0.5 [(5 xx 2^(3//2))^2 - 0] = 50 J`