Calculating Potential Energy of a Spring with a Hanging Mass

AI Thread Summary
A force of 790 N stretches a spring by 0.135 m, leading to a calculated spring constant (k) of 5851.85 N/m. The potential energy of the spring is determined using the formula U = 0.5(k)(x)^2, where the new displacement (x) is found to be approximately 0.104 m. Newton's second law can also be applied to find the displacement caused by the hanging mass. The total potential energy of the spring is ultimately calculated to be 31.5 J. This discussion highlights the importance of correctly calculating k and understanding the relationship between force, displacement, and potential energy in spring systems.
emeraldempres
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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.
 
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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?
 
What is 790/.135? Your answer for k, just by estimating in my head, is off.
 
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
 
emeraldempres said:
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}
since you've already know 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.
 
  • #10
konthelion said:
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}
since you've already know 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
 
  • #11
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.
 
  • #12
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
 
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