Question

At time $t=0$ a particle starts moving along the x-axis. If its kinetic energy increases uniformly with $t$ the net force acting on it must be
A. Constant
B. Proportional to $t$
C. Inversely proportional to ${{t}^{2}}$
D. Proportional to $\dfrac{1}{\sqrt{t}}$

Answer 1

The kinetic energy of an object is directly proportional to the mass of object and the square of its velocity. By the given relation between kinetic energy and the time, we can calculate the relation between velocity and time. Acceleration of a particle is given as the rate of change of its velocity and force is the product of mass and acceleration of the particle.

Formula used:
Kinetic energy, $K.E=\dfrac{1}{2}m{{v}^{2}}$
Acceleration, $a=\dfrac{dv}{dt}$
Force, $F=ma$

Complete step by step answer:
We are given that at time $t=0$ a particle starts moving along the x-axis with its kinetic energy increasing uniformly with $t$ and we have to calculate the net force acting on it.
The particle undergoes a one-dimensional motion, with its kinetic energy increasing uniformly with time$t$.
We know that kinetic energy of an object is formulated as:
$K=\dfrac{1}{2}m{{v}^{2}}$
Where
$m$is the mass of the object
$v$is its velocity.
So, for the particle, we have
$K\propto t$
We will remove the proportionality sign, by substituting proportionality constant $A$
So, $K=At$
Consider the particle having mass $m$and to be travelling at velocity $v$
So, for the particle, $K=\dfrac{1}{2}m{{v}^{2}}$
Or $K=\dfrac{1}{2}m{{v}^{2}}=At$
$\dfrac{1}{2}m{{v}^{2}}=At$
Or, ${{v}^{2}}=\dfrac{2At}{m}$
As, mass of the particle remains constant,
${{v}^{2}}=ct$
Where $c=\dfrac{2A}{m}$ is a constant.
Or, $v=\sqrt{ct}$

Now, to get the force, we need acceleration, as
$F=ma$
Where
$m$ is the mass
$a$ is the acceleration.
Now, we know that acceleration is equal to rate of change of velocity with time,
$a=\dfrac{dv}{dt}$
So differentiating $v=\sqrt{ct}$ with respect to $t$ ,

$\begin{aligned} & \dfrac{d}{dt}\left( v \right)=\dfrac{d}{dt}\left( \sqrt{ct} \right) \\\ & \dfrac{dv}{dt}=\sqrt{c}\dfrac{d}{dt}\left( \sqrt{t} \right) \\\ & \dfrac{dv}{dt}=\sqrt{c}\dfrac{d}{dt}\left( {{t}^{\dfrac{1}{2}}} \right) \\\ \end{aligned}$

Now, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
So, $\dfrac{d}{dt}\left( {{t}^{\dfrac{1}{2}}} \right)=\dfrac{1}{2}{{t}^{\left( \dfrac{1}{2}-1 \right)}}=\dfrac{1}{2}{{t}^{-\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{t}}$

Or, $\dfrac{dv}{dt}=\sqrt{c}\dfrac{d}{dt}\left( {{t}^{\dfrac{1}{2}}} \right)=\dfrac{\sqrt{c}}{2\sqrt{t}}$
Substituting $a=\dfrac{dv}{dt}$

We have,
$a=\dfrac{\sqrt{c}}{2\sqrt{t}}$
Or $ma=\dfrac{\sqrt{c}}{2\sqrt{t}}$

Substituting $F=ma$ we have
$F=\dfrac{\sqrt{c}}{2\sqrt{t}}$
So the force applied on the particle is equal to $\dfrac{\sqrt{c}}{2\sqrt{t}}$
Now, as $c$ is a constant,
Thus, $F\propto \dfrac{1}{\sqrt{t}}$
The force applied on the particle is inversely proportional to square root of time, $F\propto \dfrac{1}{\sqrt{t}}$
Hence, the correct answer is option D.

Note:
Kinetic energy of an object is a physical quantity that is a function of its velocity. When the velocity of an object doubles, its kinetic energy becomes four times the initial kinetic energy. Students should note that the motion $F\propto \dfrac{1}{\sqrt{t}}$ is not an equation for a standard motion. The particle is undergoing an imaginary line of motion.