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Find the potential, kinetic, and mechanical energy

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A 1.40 kg block slides with a speed of 0.885 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 637 N/m . The block comes to rest after compressing the spring 4.15 cm.

    Part A
    Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 0 cm.
    Part B
    Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 1.00 cm.
    Part C
    Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 2.00 cm.
    Part D
    Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 3.00 cm.
    Part E
    Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 4.00 cm.
    2. Relevant equations
    Potential energy= 1/2Kx2
    Kinetic energy=1/2mv2
    Mechanical energy=KE+PE

    3. The attempt at a solution
    I got the first answer correct using these equations. However, when I use these equations for part B I get them wrong. For example, this is what I did.
    part B) PE=(.5)(.01m)(.0415m)2 =8.6x10-6J
    KE= (.5)(1.40kg)(.885m/s)2 = .548 J
    ME= .548J+8.6x10-6 = .548 J
    I don't know what I am doing wrong if someone can please explain I would appreciate it! Also, I have seen in other examples that the KE is supposed to change with each compression. How? if the mass and the velocity remain the same?
     
  2. jcsd
  3. Nov 16, 2016 #2

    Doc Al

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    The velocity does not remain the same!

    As the spring is compressed, some of the block's KE transforms into the spring's PE. (Plug in the given compression for B, not .0415m.)
     
  4. Nov 16, 2016 #3

    TSny

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    Why the factor of .01 m here. If you combine the units on the on the right-hand side, do you get the correct units for energy?

    As Doc Al points out, you have not use the correct amount of compression for the spring here.
     
  5. Nov 16, 2016 #4
    I thought it had to be in meters not cm.
     
  6. Nov 16, 2016 #5

    TSny

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    You know the formula for potential energy of a spring is (1/2)kx2. It looks like you used .01 m for k.
     
  7. Nov 16, 2016 #6
    I'm just confused about it all! It looks like I am pluggining in the wrong numbers!
     
  8. Nov 16, 2016 #7
    isn't that the compression of 1cm?
     
  9. Nov 16, 2016 #8

    TSny

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    Yes. What does k stand for? What is the value of k as given in the problem?
    What does x stand for? What is the value of x for part B?
     
  10. Nov 16, 2016 #9
    Can you please explain further.....
     
  11. Nov 16, 2016 #10
    I thought x represented distance, so I assumed it was the 4.15 cm in the beginning of the problem. I thought K was the value that was given for the compression.
     
  12. Nov 16, 2016 #11

    TSny

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    You need to review the formula for PE of a spring in your class notes or your textbook. The meaning of the symbols should be explained there.
     
  13. Nov 16, 2016 #12

    Doc Al

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    That 4.15 cm is the compression when the block comes to rest (momentarily).

    K stands for the spring constant. (What is called "force constant" in the problem statement.)
     
  14. Nov 16, 2016 #13
    I just reviewed it in my textbook and I was so wrong with the units.
    Going back to part B I got,

    PE: (.5)(637N/m)(.01m)2=.03185J

    I still don't understand how the velocity changes with each compression. would I solve for a new velocity and use the ΔKE to find the new velocity?

    ΔKE= 1/2mvf2-1/2mvi2
     
  15. Nov 16, 2016 #14

    TSny

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    That looks good!

    Think about the total energy E of the system when the spring is compressed .01 m. How does that total energy compare to the initial total energy E in part A?
     
  16. Nov 16, 2016 #15
    I'm not sure :(
     
  17. Nov 16, 2016 #16

    TSny

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    As the block compresses the spring, what happens to the PE of the spring? Does it increase, decrease, or remain constant?
    As the block compresses the spring, what happens to the KE of the block? Does it increase, decrease or remain constant?
    What happens to the total energy, E? Does it increase, decrease, or remain constant (conserved)?
     
  18. Nov 16, 2016 #17
    wouldn't the potential energy increase and kinetic energy decrease? the total energy should remain conserved.
     
  19. Nov 16, 2016 #18

    TSny

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    Yes. Use the fact that the total energy is conserved to help find the KE for part B.
     
  20. Nov 16, 2016 #19
    I would use total energy=PE+KE ?
     
  21. Nov 16, 2016 #20

    TSny

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    Yes.
     
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