Under the action of force `2 kg` body moves such that its position 'x' varies as a function of time `t` given by : `x = t^(2)//2`. The work done by the force in the first `5` seconds is

Options

A. 2.5 J
B. `0.25 J`
C. 25 J
D. 250 J

Answer

C

Explanation

`W = (1)/(2) mv^(2)` where `v = (dx)/(dt)`.