Question

Under the action of force, a 2 kg body moves such that x is a function of time given by $x=t^3/3$, x in metre, t in seconds, find the work done in first two seconds.

Answer 1

Given,

x = t³/3

Differetiating x with respect to time we get,

dx/dt = v = t²

Differetiating again with respect to time we get,

a = dv/dt = 2t

For a small displacement dx, the Work done is,

dW = F. dx = ma. dx = (2)(2t)(t² dt) = 4t³dt

So, the work done in the first two seconds is,

Integrate on both the sides,

$W={\int }_0^{2}4t^{3}dt={\left[t^4\right]}_0^2=[2^4-0]=16J.$

=> W = 16 J.