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Work and Spring Energy

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A low friction cart has a spring plunger. The plunger and spring are compressed 6.80 cm and locked in place. The cart is launched from rest up an inclined track tipped at an angle of 13.5°. When the spring is released the cart travels 76.4 cm up the incline from its start position before rolling back down. (a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g? (b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm. Give reasons for your answer. (c) Describe all the energy transformations that take place from start to finish.


    2. Relevant equations

    W = ΔK + ΔU_g + ΔU_s
    K = F/Δx
    1/2mv²
    1/2kx²

    3. The attempt at a solution

    (a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g?

    So I started by using the work equation...we're assuming this system is frictionless so W = 0

    there is no kinetic energy in this one so

    1/2kx² = mgh

    Since I'm solving for k I rearranged this like

    k = (2mgh)/x²

    the thing is....when I solved for h (height) I did the .764m * sin(13.5) ......but wouldn't that be what "x" is as well?

    I know that when using

    K = F / Δx ...this Δx would be the amount the spring is compressed here....but in that other equation I think it's the distance traveled


    I'm just getting really confused here

    any hints?
     
  2. jcsd
  3. Apr 10, 2013 #2
    or actually...wouldn't "x" be the sum of the distance and the amount the spring is compressed? Because isn't "x" the distance from where the spring is relaxed?
     
  4. Apr 10, 2013 #3

    Doc Al

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    Staff: Mentor

    In this case, x is just the amount the spring was initially compressed. It's given. Not directly related to the distance traveled up the incline.
     
  5. Apr 10, 2013 #4
    That's not clicking in my head....is there a reasoning behind that? Because every other time I've used that equation, the "x" was used as the distance. Is it because of the angle and the fact that there is another energy (Gravitational P.E) in the equation?

    But okay so then it would be

    K = 2(.178N) / .0680m = 5.25 N/m?
     
  6. Apr 10, 2013 #5

    Doc Al

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    Staff: Mentor

    Realize that the cart is not attached to the spring. Once the spring uncompresses, the cart will keep on going without it.

    Double check that calculation.
     
  7. Apr 10, 2013 #6
    Oh, that makes sense now!

    Well what I did was I actually got .178352258 for mgh...my professor taught us to write down the answers with significant figures but use the original calculation in equations ...so I did

    K = 2(.178352258N) / .0680m

    which gave me the 5.25N/m
     
  8. Apr 10, 2013 #7
    Nvm sorry about that...only used the height in that mistake...so now adding in the mg portion...

    K = 2(.4299716236)/.0680

    K = 6.32N/m look better?
     
  9. Apr 10, 2013 #8
    So now

    (b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm.

    While part (a) says assume frictionless conditions....the original question says "a low friction cart has a spring plunger" ...if the system is completely frictionless then the spring will compress more than it did with just the cart resting on the spring correct? Because now the fact that Kinetic energy would have played a factor in the system with some velocity...the spring would be compressed more. ?

    (c) Describe all the energy transformations that take place from start to finish.
    Well it goes from Spring P.E in the very beginning because all the energy is in the spring, to Kinetic Energy because the object will be moving, then it will go to Gravitational Potential Energy because the system will no longer be moving (at a point) and the cart is not connected to the spring...is this correct?
     
  10. Apr 10, 2013 #9

    Doc Al

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    Staff: Mentor

    Try it one more time. Go back to your original equation.
     
  11. Apr 10, 2013 #10
    argh x²!!

    K = 2(.4299716236)/(.0680²)

    K = 92.99N/m
     
  12. Apr 10, 2013 #11

    Doc Al

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    Redo that arithmetic once more!
     
  13. Apr 10, 2013 #12
    o_O

    Okay

    K = (2mgh)/x²

    mg = 2.4108
    h = .764sin(13.5)

    mgh = .4299716236
    2mgh = .8599432472


    ***forgot the (*2) part***

    x = .0680
    x² = .004624

    (2mgh) / x²

    .8599432472 / .004624

    185.97N/m

    Sorry for those terrible mistakes!
     
  14. Apr 10, 2013 #13

    Doc Al

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    OK, now you've got it.
     
  15. Apr 10, 2013 #14

    Doc Al

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    Think about it. Whatever energy the spring gives the cart is exactly the energy needed to compress the spring.

    Not bad. Let me tweak it a bit.

    The gravitational PE increases continuously as the cart rises. The KE increases, then decreases.
     
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