Question #0eef1

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Question #0eef1

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Answer 1 · 2016-05-01T12:41:08

$9 \times U$ $9 times U$

Explanation:

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

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

And for the second load

$U_{2} = \frac{1}{2} k \cdot 3^{2} = \frac{1}{2} k \cdot 9$ $U_2=1/2k*3^2=1/2k*9$

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

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