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InterviewSolution
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A particle of mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m-1/2s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m? |
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Answer» Given m = 0.5 kg. V = ax3/2, a = 5m-1/2 s-1 initial velocity at x = 0, v1= a × 0 = 0 Final velocity at x = 2n is, V2 = a (2)3/3 = 5 × 23/2 work done = change in KE \(\frac {1}{2}\)m (v22 – v12) = \(\frac {1}{2}\)× 0.5 × 25 × 23 = 50 J . |
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