Q20  A body of mass 0.5 kg travels in a straight line with velocity $v = ax ^{3/2}$ where$a = 5 m ^{-1/2 }s ^{-1}$. What is the work done by the net force during its displacement from x = 0 to  x = 2 m ?

D Devendra Khairwa

The relation between work done and the kinetic energy is given by :

$Work\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2$

Using the relation   $v = ax ^{3/2}$  we can write :

Initial velocity  =   0    (at x  =  0 )

And          the final velocity    =   $10\sqrt{2}\ m/s$   (at  x  = 2).

Thus work done is :

$Work\ =\ \frac{1}{2}m(v^2\ -\ u^2)$

or                                                 $=\ \frac{1}{2}\times 0.5\times (10\sqrt{2})^2$

or                                                  $=\ 50\ J$

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