A vertically hanging spring is elongated by x (in equilibrium) when mass m is hanged from it. What is the work done by a man in slowly lowering the mass by a distance y further.

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Rahul Yadav · Answer

The difference of energy of two positions is equal to work done i.e 1/2 kx^ 2 - (1/2ky^2-mgy)=k/2(x^2-y^2)+mgy

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ADNAN MUHAMMED

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A vertically hanging spring is elongated by x (in equilibrium) when mass m is hanged from it. What is the work done by a man in slowly lowering the mass by a distance y further.

A vertically hanging spring is elongated by x (in equilibrium) when mass m is hanged from it. What is the work done by a man in slowly lowering the mass by a distance y further.

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A vertically hanging spring is elongated by x (in equilibrium) when mass m is hanged from it. What is the work done by a man in slowly lowering the mass by a distance y further.

A vertically hanging spring is elongated by x (in equilibrium) when mass m is hanged from it. What is the work done by a man in slowly lowering the mass by a distance y further.

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

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