Question

The relation between displacement $x$ and time $t$ for a body of mass $2Kg$ moving under the action of a force is given by $x=\frac{t^3}{3}$, where $x$ is in meter and $t$ in second, calculate the work done by the body in first $2$ seconds.

Answer 1

Given that,

Mass $M=2kg$

Displacement = $x$

Time $t=2s$

Now, the displacement is

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

On differentiate

$\frac{dx}{dt}=\frac{3t^2}{3}dt$

$v=t^{2}dt$

Again differentiate

$\frac{dv}{dt}=2tdt$

$a=2tdt$

Now, the force is

$F=ma$

$F=2\times 2t$

$F=4t$

Now, the work done is

$dW=F\cdot dx$

$dW=4t\times t^{2}dt$

$dW=4t^{3}dt$

Work done during t = 0 to 2 sec is found by integration

$\int dW=\underset{0}{\overset{2}{\int }}4t^{3}dt$

$W=[{\left(2\right)}^4-0]$

$W=16J$

Hence, the work done is $16J$