Question

A small object of mass of $100gm$ moves in a circular path. At a given instant velocity of the object is $10\hat{i}m/s$ and acceleration is $(20\hat{i}+10\hat{j})m/s^2$. At this instant of time, rate of change of kinetic energy of the object is:

• A. $200kg{m}^2/s^3$
• B. $300kg{m}^2/s^3$
• C. $10000kg{m}^2/s^3$
• D. $20kg{m}^2/s^3$

Answer 1

The correct option is D $20kg{m}^2/s^3$

Given, mass of object $\left(m\right)=100g$
$=100\times {10}^{-3}kg$
Velocity of object $\left(v\right)=10\hat{i}m/s$
Acceleration of object $\left(a\right)=(20\hat{i}+10\hat{j})m/s^2$
We know that,
$\frac{d\left(KE\right)}{dt}=F\cdot v=ma\cdot v[\because K\cdot E=1/2m{v}^2]$
$=(100\times {10}^{-3})(20\hat{i}+10\hat{j}).\left(10\hat{i}\right)$
$=100\times {10}^{-3}\times 200$
$=\frac{100\times 200}{1000}=20kg{m}^2/s^3$.