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

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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.

Jun 16th, 2015

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

Jun 16th, 2015

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Jun 16th, 2015
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Jun 16th, 2015
Dec 11th, 2016
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