Question

A ball of mass $100\text{g}$ is under a force which causes a displacement $S$ in metres, given by the relation $S=t^3+t^2$ where $t$ is in seconds. Work done by the force in 3 seconds

• A. $72.45\text{J}$
• B. $36.55\text{J}$
• C. $20\text{J}$
• D. $54.45\text{J}$

Answer 1

The correct option is D $54.45\text{J}$

Given displacement of the ball, $S=(t^3+t^2)\text{m}$........(1)
$\Rightarrow dS=(3t^2+2t)dt$
Acceleration $a=\frac{d^{2}S}{d{t}^2}=6t+2$
$\therefore$ Force acting on ball at time $t$, $F\left(t\right)=ma=0.1(6t+2)$

Work done by force at time $t$ is given by
$dW=F\left(t\right)dS=0.1(6t+2)(3t^2+2t)dt$
$=0.1(18t^3+18t^2+4t)dt$

Therefore, total work done by force in 3 seconds
$W=\int dW={\int }_0^{3}0.1(18t^3+18t^2+4t)dt$
$=0.1{[\frac{9}{2}{t}^4+6t^3+2t^2]}_0^3$
$=0.1[\frac{729}{2}+162+18]$
$=0.1\times \frac{1089}{2}=54.45\text{J}$