Question

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 ?

Answer 1

Mass of the body, m = 0.5 kg 
Velocity of the body is governed by the equation, v = a x$^{3/2}$ with a = 5 m $^{-1/2}$ s - 1
Initial velocity, u (at x = 0) = 0 
Final velocity v (at x = 2 m) = $10\sqrt{2}{m/s}$
Work done, W = Change in kinetic energy = $\frac{1}{2}$ m(v2 - u2)
= $\frac{1}{2}$× [$(10\sqrt{2})^2$ - (0)$^2$]
= $\frac{1}{2}$ × 0.5 × 10 × 10 × 2
= 50 J