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Work, Power and Energy

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Question : 20 of 30
Marks: +1, -0
A body of mass 0.5 kg travels in a straight line with velocity v=ax3/2v = a x^{3/2} where a=5m−1/2s−1a = 5 \mathrm{m}^{-1/2} \mathrm{s}^{-1}. What is the work done by the net force during its displacement from x=0→x=2x = 0 \to x = 2 m?
Solution:  
Here, Mass of body = 0.5 kg
Velocity v=ax3/2v = a x^{3/2} where a=5m−1/2s−1a = 5 \mathrm{m}^{-1/2} \mathrm{s}^{-1}
Initial velocity at x=0,v1=5×0=0x = 0, v_{1} = 5 \times 0 = 0
Final velocity at x=2,v2=5×23/2x = 2, v_{2} = 5 \times 2^{3/2}
Work done = Increase in kinetic energy =12×0.5[(5×23/2)2−0]=50J= \frac{1}{2} \times 0.5 \left[ (5 \times 2^{3/2})^{2} - 0 \right] = 50 \mathrm{J}
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