Work Energy And Power Question 137
Question: A particle of mass $ m $ moves with a variable velocity $ v $ , which changes with distance covered $ x $ along a straight line as $ v=k\sqrt{x} $ , where A: is a positive constant. The work done by all the forces acting on the particle, during the first $ t $ seconds is
Options:
A) $ \frac{mk^{4}}{t^{2}} $
B) $ \frac{mk^{4}t^{2}}{4} $
C) $ \frac{mk^{4}t^{2}}{8} $
D) $ \frac{mk^{4}t^{2}}{16} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ v=k\sqrt{x} $ Or $ \frac{dx}{dt}=k\sqrt{x} $
or $ {x^{\frac{1}{2}}}dx=kdt $
Integrating both sides, we get $ \frac{{x^{\frac{1}{2}}}}{\frac{1}{2}}=kt+C; $
Assuming $ x(0)=0 $ Therefore, $ C=0 $
$ 2\sqrt{x}=kt\Rightarrow x=\frac{k^{2}t^{2}}{4} $
or $ v=\frac{k^{2}t}{2} $
Therefore, work done, $ \Delta W= $ Increase in KE $ =\frac{1}{2}mv^{2}-\frac{1}{2}m{{(0)}^{2}}=\frac{1}{2}m{{[ \frac{k^{2}t}{2} ]}^{2}}=\frac{1}{8}mk^{4}t^{2} $
