MatthewH528275 MatthewH528275
  • 25-10-2022
  • Physics
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An elastic spring stores 45 J of potential energy when it is stretched by 2 cm. What is the spring constant?

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EbenezerV225857 EbenezerV225857
  • 25-10-2022

PE = 45 J

x = 2 cm = 0.02 m

U= 1/2 k ( x)^2

Where:

U= energy stored

x = stretch

k = spring constant

Isolate k:

k = 2U/ x^2

k = ( 2 * 45 J) / (0.02)^2

k= 225,000 N/m

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