Physics Help Forum Is this spring elastic problem erroneous?
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 Jun 30th 2014, 02:24 PM #1 Junior Member   Join Date: Jun 2014 Posts: 22 Is this spring elastic problem erroneous? 1. Elastic potential energy in a spring is directly proportional to the square of the displacement of one end of the spring from its rest position while the other end remains fixed. If the elastic potential energy in the spring is 100J when it is compressed to half its rest length, what is its energy when it is compressed to one fourth its rest length. The answer choices were 50, 150, 200, and 225J. The answer is 225J. I believe that this is wrong because the more compressed the spring is, the more potential energy it would have according to the equation PE = .5kx^2. (where x is the amount of compression). Is the word problem erroneous?
Jun 30th 2014, 08:27 PM   #2
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 Originally Posted by jlyu002 1. Elastic potential energy in a spring is directly proportional to the square of the displacement of one end of the spring from its rest position while the other end remains fixed. If the elastic potential energy in the spring is 100J when it is compressed to half its rest length, what is its energy when it is compressed to one fourth its rest length. The answer choices were 50, 150, 200, and 225J. The answer is 225J. I believe that this is wrong because the more compressed the spring is, the more potential energy it would have according to the equation PE = .5kx^2. (where x is the amount of compression). Is the word problem erroneous?
PE_0 = (1/2)k(L/2)^2 = (1/8)kL^2 = 100

PE_ = (1/2)k(L/4)^2 = (1/32)kL^2 = ?

Well we know that kL^2 = 8(100) from the first equation....

-Dan
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 Jul 1st 2014, 04:47 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The answer of 225J is correct. The trick here is to recognize that the 'x' in the equation PE=(1/2)kx^2 is not length of the spring when compressed, but rather the change of length from equilibrium. In the first case the spring is compressed to 1/2 its original length, then in the second case it is compressed to 1/4 its original length, which means it's compressed 3/2 as much as in the first case. If you want to put some numbers on it, suppose the uncompressed length is 100cm. In the first case it is compressed to 50 cm, so x = 100-50=50cm. In the second case it's compressed to 25 cm. so x = 100-25=75 cm, which is 3/2 of 50. Since PE is proportional to the amount of compression squared, the energy is 100(3/2)^2 = 225J. Last edited by ChipB; Jul 1st 2014 at 06:14 AM.
 Jul 1st 2014, 03:02 PM #4 Junior Member   Join Date: Jun 2014 Posts: 22 Thank you very much for all the generous help PhysicsForum community! However, I am still a bit skeptical with the proportions and given statements in the problem. I do not believe that the proportion is correct. X, to my understanding is the amount displaced from rest or equilibrium, or amount of compression relative to equilibrium or rest. If we imagine a spring that is 100cm, for the 1/2 length of the spring at equilibrium situation, the compression displacement would be 50. For the 1/4 length of the spring, the displacement would be 25 from equilibrium. Without plugging anything back into the (100-x)^2 formula and plugging it into the regular equation, PE=1/2kx^2, I find that the potential energy should be more for the 1/2 compressed than the 1/4 compressed. Most likely, I think I am wrong, but when I draw out the picture and compress 1/2 and compare to the 1/4 compress, I am drawn to the conclusion that the 1/2 should have more potential energy. Lastly, if we draw the spring, is equilibrium or rest position from 0cm, or from 100cm? Please, if you can correct my view on this, I would greatly appreciate it! Sincerely, Jonathan Last edited by jlyu002; Jul 1st 2014 at 03:54 PM.
 Jul 1st 2014, 04:27 PM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The problem says: "it is compressed to one fourth its rest length." If its rest length is 100cm, then 1/4 of its rest length is 25 cm. Hence it is compressed from 100 cm by an amount of 75 cm to end up at "one fourth its rest length." Hence its displacement is 75 cm. See the attached figure. Perhaps the confusion is that you are interpreting the problem statement as "it is compressed by one fourth its rest length." But that would be an incorrect interpretation. Attached Thumbnails
 Jul 1st 2014, 06:12 PM #6 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,667 I missed it myself. Thanks for the catch. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Jul 1st 2014, 07:37 PM #7 Junior Member   Join Date: Jun 2014 Posts: 22 GAhhh!!! The wording in physics always gets me. Thank you so much! You definitely blessed my psychology. This really helped! Sincerely, Jonathan
 Aug 26th 2015, 09:26 AM #8 Junior Member   Join Date: Aug 2015 Posts: 3 Ok, I get up to the point where you multiply the displacement by 1.5 in order to get 75cm. BUT WHY ARE YOU MULTIPLYING "E" WHICH IS "100" BY (1.5)^2??
 Aug 26th 2015, 09:36 AM #9 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 Because PE is proportional to displacement squared. For example if I double the displacement from equilibrium, then PE goes up by a factor of 2^2 = 4. For the case here, where displacement is increased by a factor of 1.5, PE goes up by 1.5^2 = 2.25.
 Aug 26th 2015, 12:06 PM #10 Junior Member   Join Date: Aug 2015 Posts: 3 Oh, that makes more sense now. I guess I've been overthinking this problem.

 Tags elastic, erroneous, problem, spring

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