JEE Main On 16 April 2018 Question 7
Question: A body of mass starts moving from rest along x-axis so that its velocity varies as $ v=a\sqrt{s} $ where is a constant and is the distance covered by the body. The total work done by all the forces acting on the body in the first seconds after the start of the motion is [JEE Main 16-4-2018]
Options:
A) $ \frac{1}{8}ma^{4}t^{2} $
B) $ 4ma^{4}t^{2} $
C) $ 8ma^{4}t^{2} $
D) $ \frac{1}{4}ma^{4}t^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Velocity of body $ v=a\sqrt{s} $
acceleration $ a’=\frac{dv}{dt}=\frac{a}{2\sqrt{s}}\frac{ds}{dt}=\frac{a^{2}}{2} $
displacement $ s’=\frac{1}{2}a’t^{2}=\frac{1}{2}\frac{a^{2}}{2}t^{2} $
Work done $ W=force\times displacement $
$ W=ma’\times s’=\frac{1}{8}ma^{4}t^{2} $
