Question

A particle of mass 0.5 kg travels in a straight line with velocity v = ax^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?

Answer 1

Here m = 0.5 kg

v = ax^3/2, a = 5 m^-1/2 s^-1

Initial velocity at x = 0, v₁ = a x 0 = 0

Final velocity at x = 2, v₂ = a(2)^3/2 = 5 x (2)^3/2

Work done = Increase in K.E.

= 1/2 m(v₂² - v₁²)

= 1/2 x 0.5 [(5 x 2^3/2)² - 0] = 50 J