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An automobile spring extends 0.2 m for 5000 N.The ratio of potiential energy stored in the spring when it has been compressed by 0.2 m to the potiential energy stored in a 10 micro farad capacitor at a potiential difference of 10,000 volt will be?

Aastha Aggarwal , 7 Years ago

Grade 12

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

For a spring, $F=kx\implies 5000=k(0.2)\implies k=25000N/m$

Potential energy stored in the spring= $\dfrac{1}{2}kx^2=\dfrac{1}{2}25000\times 0.2^2J=500J$

Energy stored in the given capacitor= $\dfrac{1}{2}CV^2=\dfrac{1}{2}\times (10\times 10^{-6})\times 10000^2J=500J$

Hence ratio of the energies stored=1:1

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