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  • A body of mass 0.5 kg travels in a straight line with velocity v equals a x raised to 3/2 where a equals 5 m raised to –1/2 s raised to –1. What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

 Q20  A body of mass 0.5 kg travels in a straight line with velocity v = ax ^{3/2} wherea = 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 ?

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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

Posted by

Devendra Khairwa

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