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Can't seem to get this energy problem

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data

    An ideal massless spring S can be compressed 1 metre by 100 N. That same spring is placed at the bottom of a frictionless incline which makes a 30 degree angle with the horizontal. A 10-kg block M is released from the top, and this block compresses the spring by 2 metres before coming to a rest.

    a) Find the total distance traversed by the block on the incline

    b) Find the speed of the block just before it hits the spring [ I suppose I could probably get this if I get a) ]



    2. Relevant equations

    W = K1 - K0

    E = K + U

    0 = change in K + change in U
    ... et c





    3. The attempt at a solution

    A couple of things are confusing to me.. If the spring has been compressed by 2 metres, then the block acts with a force on 200 N onto the spring.. but where does this magnitude come from? The only thing that would slide the block down is the weight.. and that isn't 200 N or anything like that.

    If I knew how to explain that, I think that I would first figure out the kinetic energy as it hits the spring and work from there - since I can solve the equation W [Fd] = K1 - K0 and K0 would just be 0.

    Thanks
     
    Last edited: Jan 3, 2010
  2. jcsd
  3. Jan 3, 2010 #2
    Elastic PE stored in compressed spring is (1/2)kx2=(1/2)(100N/m)(2m)2=200J. This is the KE of block at bottom of incline because this is the amount of energy that the block loses in compressing the spring. The block gains KE in the amount mgh as it slides down incline. mgh = 200J, h=2.04 m. d(plane length)=h/sin30 = 4.08m. Then (1/2)mv2=200J, v=6.32m/s.
     
    Last edited: Jan 3, 2010
  4. Jan 3, 2010 #3
    Thanks for the reply. I'm sorry but I can't really understand your answer :S Could you please explain it some more? Where did you get h = 2.04..et c?

    And the answer should be 4.1m for a)

    Thanks!
     
  5. Jan 3, 2010 #4
    You need use only energy conservation.

    At the bottom of the incline:
    energy lost by the block as the spring slows it to a stop (the block's kinetic energy at the bottom of the incline) = energy gained by the spring = (1/2)(100N/m)(2m)^2 = 200J

    As the block slides down the incline:
    energy at the top = energy at the bottom
    Ki + Ui = Kf + Uf
    Ui = Kf
    mgh = 200 J where h is the height of the incline (the altitude)
    (10kg)(9.81m/s^2)(h) = 200J
    h = 200/(10x9.81) m
    h = 2.04 m

    The incline is a right triangle with height h (the altitude) and hypotenuse d (the plane length). sin30degrees = h/d, then d = 4.1m.

    The kinetic energy Kf at the bottom of the incline is 200J, which is equal to (1/2)mv^2, so (1/2)(10kg)v^2=200J, then v^2=2(200)/10 m^2/s^2 = 40m^2/s^2, then v = 6.32 m/s.

    You could use the work-kinetic energy theorem to solve this problem, but the solution would be more complicated.
     
  6. Jan 3, 2010 #5
    okay, I got it! Thanks :) but the answer to the second part should be 4.5 m/s, what happened?
     
    Last edited: Jan 3, 2010
  7. Jan 3, 2010 #6

    vela

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    The component of the block's weight along the incline is 49 N. Until the spring is compressed to the point where it exerts 49 N, the net force on the block is still down along the incline, and the block will be speeding up. It's only after the spring starts exerting a force greater than 49 N does the block slow down, so when the block finally comes to rest, the spring will be exerting a force greater than 49 N, possibly much greater.

    How much the spring is ultimately compressed depends on, not only the block's weight, but on its initial speed. The spring will be compressed more if the block is moving faster initially than if it were moving slowly.
     
  8. Jan 3, 2010 #7
    Thanks, I think I understand. The spring makes the mass exert a greater force onto the spring.

    By the way, can you see what happened? the answer for part b) should be 4.5 m/s, but the answer Heidi got was 6.32 m/s :S

    Thanks
     
  9. Jan 3, 2010 #8
    Oops. When the block first hits the spring it will have gravitational potential energy as well as kinetic. That 200 J of energy is divided into 101.9 J of kinetic and 98.1 J of potential, giving a speed of 4.5 m/s.
     
  10. Jan 3, 2010 #9
    Note: when the block first hits the spring its altitude will be 1 m and its gravitational potential energy will be mgh = 10x9.81x1 J.
     
  11. Jan 3, 2010 #10
    Ah okay, thanks!
    And just to clarify a point here:


    Ki + Ui = Kf + Uf
    Ui = Kf

    Ki = 0 and Uf = 0 because we are considering the mass as it slides to some point... where in our case, our mass will have no potential energy (it will stop moving there, no more gravitational PE). and Ki = 0 because v0 will equal to zero at some point.

    THanks!
     
  12. Jan 3, 2010 #11

    vela

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    Yes, I see what happened. She made a mistake. The answer is 4.5 m/s. What did you get?

    (I guess I should refresh the thread before I post.)
     
  13. Jan 3, 2010 #12
    Okay, and you got the altitude by 1.7/sin(90) = a/sin(30) ?

    Thanks
     
  14. Jan 3, 2010 #13
    Ki is zero because the problem states that the block is "released." So you assume it starts from rest. You make the choice to set the zero level for gravitational potential energy at the block's lowest altitude (you don't have to), so Uf will be zero.
     
    Last edited: Jan 3, 2010
  15. Jan 3, 2010 #14
    (2m)(sin30)=1m.

    2m (the compression distance) is the hypotenuse.
     
  16. Jan 3, 2010 #15
    excellent, it's all clear to me now! Thanks!

    By the way, I appreciate all the help.. this is the first conservation of energy question I've done so sorry for dragging it out for so long
     
  17. Jan 4, 2010 #16
    Of course it was dragged out for so long...because of my mistake!

    Glad you've got it all figured out.
     
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