Question

A force acts on a $3\text{gm}$ particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3,$ where $x$ is in metres and $t$ is in seconds. The work done during the first $4$ second is

• A. $576\text{mJ}$
• B. $450\text{mJ}$
• C. $490\text{mJ}$
• D. $528\text{mJ}$

Answer 1

The correct option is D $528\text{mJ}$

Displacement of the particle as a function of time is given by
$x=3t-4t^2+t^3\frac{dx}{dt}=3-8t+3t^2$
$dx=(3-8t+3t^2)dt$
acceleration $\Rightarrow a=\frac{d^{2}x}{d{t}^2}=-8+6t$

Work done by variable force is $W=\int Fdx$
$\Rightarrow W=\int madx$
$\Rightarrow W=m\int (-8+6t)(3-8t+3t^2)dt$
$\Rightarrow W=m{\int }_0^t(18t^3-72t^2+82t-24)dt$
$\Rightarrow W=m{[18\frac{t^4}{4}-72\frac{t^3}{3}+41t^2-24t]}_0^4$
$\Rightarrow W=m[18\frac{4^4}{4}-72\frac{4^3}{3}+41(4{)}^2-24\left(4\right)]$
$\Rightarrow W=0.003\left(176\right)=0.528J=528mJ$