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A body of mass 0.5 kg travels in a straight line with velocity u= a x^3/2 where a= 5m^-1/2s^-1. The work done by the net force during its displacement from x = 0 to x = 2m is :

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1.5 J

50 J

10 J

100 J

answer is B.

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Detailed Solution

here initial velocity is zero and final velocity $v = 5 {(2)}^{3 / 2}$

work done = change in kinetic energy

$\frac{1}{2} m v^{2} = \frac{1}{2} (0.5) {[5 (2^{\frac{3}{2}})]}^{2} = 0.25 \times 25 \times 2^{3} = 50 J$

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