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Potential Energy of a Spring

  1. Feb 15, 2007 #1
    New member here...

    I have the following homework problem:
    "A 200-g mass attached to the end of a spring causes it to stretch 5.0 cm. If another 200-gmass is added to the spring, the potential energy of the spring will be"

    I know the answer is 4 times as much but I am not quite sure how the answer is being reached.

    I thought I could use the forumla:
    U= 1/2 kx^2

    But I am wrong.

    Any help would be much appreciated!
    Thank you!!!
    A.
     
  2. jcsd
  3. Feb 15, 2007 #2

    Kurdt

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    You have the correct formula for the potential of a spring. Are you not supposed to find the final potential of the spring once the other mass is added. Tell us exactly what you've tried.

    P.S. Welcome to the forums.
     
  4. Feb 15, 2007 #3
    Maybe, I did this wrong, but...

    I doubled up the two 200s...

    Then I took the spring constant:
    K=400/5
    K=20

    Then I used the potential energy formula:
    U=1/2(20)(5)^2
    and I came up with 250

    Which is probably ALL wrong.
     
  5. Feb 15, 2007 #4

    Kurdt

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    Your calculation of the spring constant is wrong. What you need to do is use the original data for 200 grams to get the spring constant. The force exerted on the spring by the mass is given by its weight (i.e. weight = mg). And then you should use the equation that relates the force exerted on a spring to the distance moved (F=-kx). Remember to use SI units aswell.
     
  6. Feb 15, 2007 #5
    Could you...

    You first need to find the Spring Constant:
    K=f/x=200/5=40

    So your spring constant (k) is 40, your displacement (x) is the 5cm.

    Now, first find your potential energy without the extra 200 grams added…
    u=1/2kx^2=1/2(40)5^2= 500

    Now, find your second potential energy with the added 200 grams added… Since you’ve doubled the weight (k will remain the same) BUT your displacement will increase to 10 (as you doubled).
    u= kx =1/2(40)10^2 = 2,000

    So, now… 2000 divided by 500 equals “4 times as much”.
     
  7. Feb 15, 2007 #6

    Kurdt

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    Well it is going to be 4 times as much but what you're doing is wrong. You can't say k =200/5 because 200 is a mass not a force. The force you want to use is the weight of the 200g mass. Weight as I said above is the mass multiplied by the gravitational acceleration (9.8 m/s2).

    What the question is asking is what the potential energy will be so you want the actual value. Also can I stress again please use SI units its very much safer. So instead of grams convert to Kilograms and instead of centimeters use meters.

    Other than that your method is correct.
     
  8. Feb 15, 2007 #7

    JK423

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    alemon,
    Why dont you leave the numbers and just play with the symbols? Its much much easier and certainly not confusing at all.
    You dont get confused with the SI units as well ..
    Try working that way. Only subsitute the numbers at the last formula.
     
  9. Oct 17, 2008 #8
    why do we assume that the string's displacement will double?
     
  10. Oct 17, 2008 #9

    Doc Al

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    You don't assume it, it's a consequence of Hooke's law.
     
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