A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential en

Physics
Tutor: None Selected Time limit: 1 Day

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case?

a.
b.
c.
d.

No answer is correct.

Jun 16th, 2015

Thank you for the opportunity to help you with your question!

First, let's remember the potential energy stored by the spring would be:

E = (1/2)(K)(X^2) ;   where: E is the potential energy (in Joules), K is the constant of the spring (in N/m) and X is the distance (in this case what the spring is compressed, in meters).

So from the above formula we solve for K like this:

E = (1/2)(K)(X^2)  ----------->  2*E = 2*(1/2)(K)(X^2)  ------------> 2E = (K)(X^2)

2E = (K)(X^2)  ---------------> 2E/X^2 = (K)(X^2)/X^2  ---------------> 2E/X^2 = K    so    K = (2E)/X^2

So it is giving X = 12.0 cm (we need to convert it into meters) and W = 72.0 J. Then we have:

X = 12.0 cm * (1 m/ 100cm) = 0.12 m  ------> X = 0.12 m

Finally, we have:

K = (2(72.0 J))/(0.12 m)^2 = 144 J / 0.0144 m^2 = 144 N.m / 0.0144 m^2 = 10000 N/m

K = 10 000 N/m  , so option A

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 16th, 2015

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