# Question #0eef1

May 1, 2016

$9 \times U$

#### Explanation:

If the spring obeys Hooke's law, then the potential energy, $U$, stored for an extension, $x$, will be
$U = \frac{1}{2} k {x}^{2}$
Where $k$ = spring constant

For the first load , $x = 1$ so
$U = \frac{1}{2} k \cdot {1}^{2} = \frac{1}{2} k$ [equation 1]

And for the second load

${U}_{2} = \frac{1}{2} k \cdot {3}^{2} = \frac{1}{2} k \cdot 9$

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