A body of mass 1 kg moves along a straight line with a velocity v =...

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: $1250$