Question #0eef1

1 Answer
May 1, 2016

9 times U9×U

Explanation:

If the spring obeys Hooke's law, then the potential energy, UU, stored for an extension, xx, will be
U=1/2kx^2U=12kx2
Where kk = spring constant

For the first load , x=1x=1 so
U=1/2k*1^2=1/2kU=12k12=12k [equation 1]

And for the second load

U_2=1/2k*3^2=1/2k*9U2=12k32=12k9

By comparing this to equation 1 we can see that the PE in terms of U for the second load is 9times U9×U