Work Energy And Power Question 315
Question: A body of mass m accelerates uniformly from rest to a speed $ ( \lambda ) $ in time $ \infty $ . The work done on the body till any time t is
Options:
A) $ \frac{1}{2}mv_0^{2}( \frac{t^{2}}{t_0^{2}} ) $
B) $ \frac{1}{2}mv_0^{2}( \frac{t_0}{t} ) $
C) $ mv_0^{2}( \frac{t}{t_0} ) $
D) $ mv_0^{2}{{( \frac{t}{t_0} )}^{3}} $
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Answer:
Correct Answer: A
Solution:
[a] $ v_0=at_0 $
$ \therefore $ $ a=\frac{v_0}{{t _{{}}}} $
Velocity at any time t would be
$ V=at=( \frac{v_0}{t_0} )t $
$ \therefore $ Kinetic energy,
$ \text{K=}\frac{1}{2}mv^{2}=\frac{1}{2}m{{( \frac{v_0}{t_0} )}^{2}}t^{2} $
From work energy theorem
$ W=K.E. $
or $ W=\frac{1}{2}mv_0^{2}( \frac{t^{2}}{t_0^{2}} ) $