A particle of mass 0.1 kg is subjected to a force that varies with distance as shown in the figure. If it starts its journey from rest at x = 0, its velocity (in m/s) at x =12m is

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answer is 40.

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Detailed Solution

Work done by the force is $W = Area of the F - x curve = (\frac{4}{2} + 4 + \frac{4}{2}) \times 10 = 80 J From work - energy theorem, W = \frac{1}{2} {mv}^{2} or v = \sqrt{\frac{2 W}{m}} = \sqrt{\frac{2 \times 80}{0 .1}} = 40 m / s$

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