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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?

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?

Grade:12

1 Answers

Eshan
askIITians Faculty 2095 Points
4 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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