A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be n times. The value of n is
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answer is 16.
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Detailed Solution
Potential energy U=12kx2∴U∝x2 [if k = constant]If elongation made 4 times then potential energy will become 16 times.
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A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be n times. The value of n is