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  • 23-02-2017
  • Physics
Answered

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?

Verified Answer :

scme1702
scme1702 scme1702
  • 01-03-2017

The elastic potential energy in a spring is given by:
EPE = 1/2 kx²
x = √(2 x 47 / 30,000)
x = 0.056 m

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