Question

A time dependent force $F=\beta t$ acts on a particle of mass $4\text{kg}$. If the particle starts from rest, the work done by the force during the first $1$ second will be [$\beta$ is a (+)ve constant]

• A. $\frac{\beta }{32}$
• B. $\frac{32}{\beta }$
• C. $\frac{{\beta }^2}{32}$
• D. $\frac{{\beta }^2}{16}$

Answer 1

The correct option is C $\frac{{\beta }^2}{32}$

Given that
$F=\beta t$
and we know
$F=ma=m\frac{dv}{dt}$
$\therefore m\frac{dv}{dt}=\beta t$
$\Rightarrow mdv=\beta tdt$
Integrating both sides,
$\underset{0}{\overset{v}{\int }}4dv=\underset{0}{\overset{1}{\int }}\beta tdt$
$\Rightarrow 4v=\beta \times \frac{1}{2}=\frac{\beta }{2}$
$\Rightarrow v=\frac{\beta }{8}$
$\therefore$ From work-energy theorem,
$W=\Delta KE=\frac{1}{2}m(v^2-u^2)=\frac{1}{2}m(v^2-0^2)=\frac{1}{2}\times 4\times \frac{{\beta }^2}{64}=\frac{{\beta }^2}{32}$
$\Rightarrow W=\frac{{\beta }^2}{32}$