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A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?

A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?
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A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?

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User SELA
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1 Answer

1 vote
The elastic potential energy in a spring is given by:
EPE = 1/2 kx²
x = √(2 x 47 / 30,000)
x = 0.056 m
answered
User Swizec Teller
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8.5k points
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