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प्रश्न
A bullet of an air gun weighs 0.01 kg. It is propelled out from the air gun with a velocity of 40 ms−1. Calculate the potential energy of the spring.
संख्यात्मक
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उत्तर
\[ \text{Kinetic energy of bullet} = \frac{1}{2}\, mv^{2} \]
\[= \frac{1}{2}\, (0.01\ \mathrm{kg})\, (40\ \mathrm{m\,s^{-1}})^{2} \]
= 8 J
By the law of conservation of energy, Potential energy of spring = Kinetic energy of bullet = 8 J.
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