Question

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If a spring was stretched to 2 times its normal length, the elastic potential energy in the spring would change by a factor of ___? Round to the nearest whole number.

Answer 1

Final elastic potential energy:
`(1)/(2)kL_0^2`

Step-by-step explanation:

The elastic potential energy of a spring is given by

`U=(1)/(2)k(L-L_0)^2`

where

k is the spring constant

`L` is the length of the spring

`L_0` is the natural length of the spring

Therefore, for a spring at its natural length,

`L=L_0`

So its elastic potential energy is zero:

`U=0`

If the spring is stretched to 2 times its normal length,

`L=2L_0`

Therefore its elastic potential energy would be

`U=(1)/(2)k(2L_0-L_0)^2=(1)/(2)kL_0^2`