Question

Seen the displacement X and time t for a body of mass 2 Kg moving $x=\frac{t^3}{3}$, where x is in meter and t in sec. Calculate the work done by body in first 2 second.

Answer 1

Given that,

Mass $m=2kg$

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

Now, the velocity is

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

On differentiating

$v=\frac{dx}{dt}$

$v=\frac{d\left(\frac{t^3}{3}\right)}{dt}$

$v=t^2$

$dx=t^{2}dt$

Again on differentiating

The acceleration is

$a=\frac{dv}{dt}$

$a=\frac{d\left(t^2\right)}{dt}$

$a=2t$

Now, the force is

$F=ma$

$F=2\times 2t$

$F=4t$

Now, the work is

$dW=F\cdot dx$

$dW=4tdx$

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

$dW=4t^{3}dt$

Now, the work done during t =0 to t= 2 s

So,

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

$W={\left[t^4\right]}_0^2$

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

$W=16J$

Hence, the work done is $16J$