Question

A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along x-axis. Its kinetic energy $K$ changes with time as $\frac{dK}{dt}=rt$ where $r$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

• A. The force applied on the particle is constant.
• B. The speed of the particle is proportional to time.
• C. The distance of the particle from the origin increases linearly with time.
• D. The force is conservative.

Answer 1

Answer: (A), (B), (D)

The force is conservative.

Given that,

$\frac{dK}{dt}=rt$
We know that, Kinetic energy $K=\frac{1}{2}m{v}^2$
Differentiating with respect to time on both sides , we get
$\frac{dK}{dt}=mv\frac{dv}{dt}$

$\Rightarrow \frac{dK}{dt}=rt=mv\frac{dv}{dt}$
$\Rightarrow mv\frac{dv}{dt}=rt$

Integrating on both sides,

${\int }_0^{v}vdv={\int }_0^t\frac{r}{m}tdt$
$\Rightarrow \frac{v^2}{2}=\frac{r{t}^2}{2m}\Rightarrow v=\sqrt{\frac{r}{m}}t.....\left(1\right)$

Differentiating with respect to time on both sides,

$a=\frac{dv}{dt}=\sqrt{\frac{r}{m}}$

From this,

Force $\left(F\right)=ma=\sqrt{rm}=\text{constant}.....\left(2\right)$

From $\left(1\right)$ we know that,

$v=\frac{ds}{dt}=\sqrt{\frac{r}{m}}t$

Integrating on both sides,

$\Rightarrow s=\sqrt{\frac{r}{m}}\frac{t^2}{2}+C$

Since $s=0$ at $t=0$, we get

$s=\sqrt{\frac{r}{m}}\frac{t^2}{2}$

Also, From $\left(2\right)$ since $F$ is constant function, it represents a conservative force.

Hence, options (a), (b) and (d) are correct.