Question

Position-time variation of a particle of mass $1kg$ is $x=(2t^3+3)m$. Calculate the work done by the force in first $2seconds$ .

Answer 1

Given, $x=2t^3+3$

$\Rightarrow dx=6t^2$

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

$\Rightarrow a=12tAnd,Force=1\times 12t=12t$

So, the word done in first two second $=\int Fdx$

$\Rightarrow W={\int }_0^{2}72t^3$

$=288J$