# Potential energy of a spring

A force of 790 N stretches a certain spring a distance of 0.135 m. What is the potential energy of the spring when a mass of 62.0 hangs vertically from it?

I thought starting with finding the spring constant k would be good and i got it to be 585.85. i then plugged it into the potential energy equation u= .5(k)(x)^2 where x is the displacement. I got the new displacement by setting up a porportional equation and the new x was .104 m.

All the answers look correct except for k.

when i calculated k this is what i did

Spring Force(790 N) = (k)(distance(.135m))
and isolated the variable.

what did i do wrong?

790 is the force in newtons and i divided it by .135 which is the distance the weight on the spring made the spring stretch, the answer was 5851.85

790 is the force in newtons and i divided it by .135 which is the distance the weight on the spring made the spring stretch, the answer was 5851.85
Ok. You wrote
585.85
Everything else is fine.

Another way to find the displacement would be to set up Newton's second law as well.

$$mg+(-k\Delta y=0) \Rightarrow \Deltay=\frac{mg}{k}$$

Then plug back into the potential energy of the spring. $$U_{s}=\frac{1}{2}ky^2$$

does that help me with finf=ding the potnetial energy?

sorry,i meant does that help me with finding the potential energy?

Yes. The potential energy of a spring is just a formula unless you want to find the total change in potential energy = gravitational potential energy + potential energy of the spring.

Another way to find the displacement would be to set up Newton's second law as well.

$$mg+(-k\Delta y=0) \Rightarrow \Deltay=\frac{mg}{k}$$

Then plug back into the potential energy of the spring. $$U_{s}=\frac{1}{2}ky^2$$

using the mg/k equation gave me 1.03 m as my displacement but that does not sound right if the force 0f 709 only gave a displacement of .135 m

Actually you would get ~.1038 =~ .104m which is the distance you're trying to find.

You've already found k, you need to find the distance y(the distance that the object stretches the spring, hence why I used Newton's 2nd law). There are two distances in this problem., one is already given, the other(the distance that the mass is stretching the string so that you can find the potential energy of the spring)is the one you're trying to find.

thank you so much for your help. i am doing my home work online and my summer school teaher gave us a homework assignment due each night of the weekend but is unavailable himself to answer questions. i got the answer though 31.5 J