Question

Question #0eef1

Answer 1

`9 times U`

Explanation:

If the spring obeys Hooke's law, then the potential energy, `U`, stored for an extension, `x`, will be
`U=1/2kx^2`
Where `k` = spring constant

For the first load , `x=1` so
`U=1/2k*1^2=1/2k` [equation 1]

And for the second load

`U_2=1/2k*3^2=1/2k*9`

By comparing this to equation 1 we can see that the PE in terms of U for the second load is `9times U`