Question

A small object of mass of $100\text{g}$ moves in a circular path. At a given instant velocity of the object is $10\hat{i}m{s}^{-1}$ and acceleration is $(20\hat{i}+10\hat{j})m{s}^{-2}$ . At this instant of time, the rate of change of kinetic energy of the object is (in SI units)

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

Answer 1

Answer: (D)

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