Question

The displacement $x$ of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by $x=\frac{t^3}{3}$ (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.

Answer 1

Given,

$mass=1kg$ and

$X=\frac{t^3}{3}$ Where x is in meter and t in second.

So,

$X=\frac{t^3}{3}$

By differentiating with respect to its derivatives.

$dx=t^{2}dt$

$\frac{dx}{dt}=v=t^2$

$\frac{d^{2}x}{d{t}^2}=a=2t$

Force acting on the body $=ma=20t$ Newtons

So work done $W={\int }_{t=0}^{t=t}Fdx$

$={\int }_0^{t}20t\times t^{2}dt$

$=\frac{20}{4}[t^4{]}_0^t$

$=5t^4$

Thus the work done in the last second is $W(t=1sec)=5Joules$