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Stretched Spring

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A spring with 58 hangs vertically next to a ruler. The end of the spring is next to the 15- mark on the ruler. If a 2.0- mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

    2. Relevant equations

    I used F=KX should it be mgy=kx? or mgy=1/2kx^2

    3. The attempt at a solution

    My solution was 32 centimeters and 34 centimeters and those were both wrong. Must use two sig figs.
  2. jcsd
  3. Oct 12, 2008 #2


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    With F=-kx you can know k by the original measurement - assuming that 0 is the 0 force point. This implies that each unit represents 58*g/15 units of force. Adding another 2*g units of force means that you will get an additional 2*g*15/58*g units. This suggest then it will lengthen to 15+(30/58) units.

    I'm curious where you got cm out of your description. Or is there a picture you aren't sharing and other units you haven't mentioned?
  4. Oct 12, 2008 #3
    oh the ruler was 15 cm
  5. Oct 12, 2008 #4
    sorry i guess my copy paste didnt work. The 58 is 58 N/m and is the K and it hangs next to a spring that is next to a ruler at 15 cm. If you add 2 kg to it what is the new measured distance?
  6. Oct 12, 2008 #5


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    I still have no idea where the origin is. But 2 kg will move it 2*9.8/58 m.
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