Question

A particle of mass $2kg$ starts moving in straight line with an initial velocity of $2m/s$ at a constant acceleration of $2m/s^2$. The numerical value of rate of change of kinetic energy at any moment is:

• A. four times the numerical value of velocity at that moment
• B. two times the numerical value of displacement at that moment
• C. four times the numerical value of rate of change of velocity at that moment
• D. constant throughout

Answer 1

The correct option is A four times the numerical value of velocity at that moment

We know that kinetic energy is given by $K.E.=\frac{1}{2}m{v}^2$

$d\left(KE\right)/dt=mv\left(\frac{dv}{dt}\right)=mva$

where $m$ is the mass

$a$ is the acceleration and

$v$ is the velocity

Given that $m=2kg,$ $a=2m/s^2$ and $v=2m/s$

Therefore, $\frac{d(K.E.)}{dt}=\left(2\right)\left(2\right)\left(v\right)=4v$

Therefore, the rate of the change of kinetic energy is four times the numerical value of velocity.