A body of mass \(1\) kg moves along a straight line with a velocity \(v = 2x^2\). The work done by the body during displacement from \(x = 0\) to \(5\) m is __________ J.

Given mass \(m = 1\) kg and velocity \(v = 2x^2\).

According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the body.

\(W = \Delta K = K_f - K_i\)

At \(x = 0\) m, initial velocity \(v_i = 2(0)^2 = 0\) m/s.

At \(x = 5\) m, final velocity \(v_f = 2(5)^2 = 50\) m/s.

Initial kinetic energy \(K_i = \dfrac{1}{2} m v_i^2 = 0\) J.

Final kinetic energy \(K_f = \dfrac{1}{2} m v_f^2 = \dfrac{1}{2} \times 1 \times (50)^2 = 1250\) J.

Work done \(W = 1250 - 0 = 1250\) J.

Answer: \(1250\)