Question

A time dependent force $F=p+q{t}^2$ acts on a particle of mass $3\text{kg}$. If the particle starts from rest, the work done by the force during the first $3$ second will be [$p$ and $q$ are $+ve$ constant]

• A. $3p^2+18pq+27q^2$
• B. $1.5p^2+18pq+27q^2$
• C. $1.5p^2+18pq+13.5q^2$
• D. $1.5p^2+9pq+13.5q^2$

Answer 1

The correct option is D $1.5p^2+9pq+13.5q^2$

Given that

$F=p+q{t}^2$

Using, $F=ma=m\frac{dv}{dt}$

$\therefore m\frac{dv}{dt}=p+q{t}^2$

$\Rightarrow mdv=(p+q{t}^2)dt$

Integrating both sides,

$\underset{0}{\overset{v}{\int }}3dv=\underset{0}{\overset{3}{\int }}(p+q{t}^2)dt$

$\Rightarrow 3v={[pt+\frac{q{t}^3}{3}]}_0^3$

$\Rightarrow v=p+3q$

$\therefore$ From work-energy theorem,

$\text{Workdone}\left(W\right)=\Delta KE=\frac{1}{2}m(v^2-u^2)=\frac{1}{2}m(v^2-0^2)$

$\Rightarrow W=\frac{1}{2}\times 3\times (p+3q{)}^2$

$\Rightarrow W=(1.5p^2+9pq+13.5q^2)$

Henec, option $\left(d\right)$ is the correct answer.