Question

Kinetic energy of a particle moving in a straight line is proportional to the time $t$. The magnitude of the force acting on the particle is:

• A. directly proportional to the speed of the particle
• B. inversely propotional to $\sqrt{t}$
• C. inversely proportional to the speed of the particle
• D. directly proportional to $\sqrt{t}$

Answer 1

Answer: (A), (B)

inversely propotional to $\sqrt{t}$
directly proportional to the speed of the particle

Given that,

$\frac{1}{2}m{v}^2\propto t$

$\frac{1}{2}m{v}^2=kt$

$v=\sqrt{\frac{2k}{m}}\times {\left(t\right)}^{\frac{1}{2}}.....\left(I\right)$

Now, the acceleration

Differentiate of equation (I)

$\frac{dv}{dt}=\sqrt{\frac{2k}{m}}\times \frac{1}{2}{\left(t\right)}^{\frac{-1}{2}}$

$\frac{dv}{dt}=\sqrt{\frac{2k}{m}}\times \frac{1}{2\sqrt{t}}$

The magnitude of the force acting on the particle is

$F=m\times \frac{dv}{dt}$

$F=m\times \sqrt{\frac{2k}{m}}\times \frac{1}{2\sqrt{t}}$

$F=\sqrt{2mk}\times \frac{1}{2\sqrt{t}}$

Also;

Acceleration, $a=\frac{1}{\sqrt{t}}\sqrt{\frac{k}{2m}}$

or

$a=\frac{v}{2t}$

Thus,

Acceleration is directly proportional to the velocity.

Hence, the magnitude of the force acting on the particle is inversely proportional to$\sqrt{t}$