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Really need help on simple energy and spring on incline

  1. Oct 21, 2007 #1
    A 2.00 kg block is placed against a spring on a frictionless 30.0° incline (Fig. 8-33). (The block is not attached to the spring.) The spring, whose spring constant is 19.6 N/cm, is compressed 16.0 cm and then released.

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    (a) What is the elastic potential energy of the compressed spring?
    So I think of it like this, it wants the answer in joules, so I convert 19.6N/cm to .196N/m, and 16cm to .16m anyway. PE=1/2 K X^2
    so
    PE=.5(.196)(.16^2) = .0025088J but it's wrong. I know this is very simple, But I don't understand why I am wrong here
     
  2. jcsd
  3. Oct 21, 2007 #2

    hage567

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    This is wrong. Try again, make sure your units work out.
     
  4. Oct 21, 2007 #3
    I tried leaving them the same, but that didn't work either. The only other thing I can think of is leaving one the same and changing the other, which is mixing units, and thats wrong. What is wrong about it, can you be a little more specific?
     
  5. Oct 21, 2007 #4

    hage567

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    You just didn't convert it properly. Write it out so you can see how the units need to be:

    19.6 N/cm * 100 cm/m = 1960 N/m.
     
  6. Oct 21, 2007 #5
    Ahh thank you, I swear I lose so many points because of stuff like this
     
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