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Displacement of box on inclined plane due to spring

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data
    A frictionless, incline plane at 20 degrees has a spring with force constant k=500N/m attached to the bottom, parallel to the inclined plane.

    A block of mass m=2.5kg is placed on the plane at a distance d =3m from the spring. From this position the block is projected downward toward the spring with speed v=0.75m/s. By what distance is the spring compressed when the block momentarily comes to rest?

    3. The attempt at a solution

    F(x)=mg+nsin(20)
    F(y)=ma+(-kx)+ncos(20)

    Do i use the velocity given to find the acceleration of the block and hence the force?
    can i do this using vf^2=vi^2+2ad where a is 9.8?
    also i have failed to even incorporate the displacement of the spring x which is what we're looking for in the first place. how do i incorporate it ?

    Please correct me on anything Ive done wrong
    Thanks!

    Haylee
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 10, 2013 #2
    Hello hkor, I think you should consider conservation of energy for this problem.

    What kind of energy do you have at the beginning (top of incline), and what kind at the end (bottom)?
     
  4. Jun 10, 2013 #3
    conservation of energy

    Do you mean mgh(initial) +0.5mv^2(initial)= mgh(final) +0.5mv^2
    where the initial velocity of the box is 0.75, the initial height .3m and the final velocity of the box zero, h final is the only unknown?
     
  5. Jun 10, 2013 #4

    gneill

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    Staff: Mentor

    There's a bit more to it; you need to include the potential energy that the spring gains as it is compressed.
     
  6. Jun 10, 2013 #5
    Please look at the response from gneill. If you would use only this formula, how would you determine spring compression?

    When we talk about h, do you know how to determine it?
     
  7. Jun 10, 2013 #6
    Okay so at the initial position the box has PE= mgh and KE = 0.5mv^2.
    At its final position the box is a rest therefore KE = 0 and all energy is transferred into the PE of the spring?

    using the elastic potential energy of a spring

    mgh + 1/2mv^2 = 1/2kx^2 + 0
    (2.5kg)(9.8N)(h) +(0.5)(2.5kg)(0.75m/s)^2 = (0.5)(500N/m)(x)^2
    where h = .3sin(20')

    Solve for x?
     
  8. Jun 10, 2013 #7

    gneill

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    Staff: Mentor

    The box continues to lose height as the spring compresses, so more gravitational potential energy is involved. You might consider working the problem from when the box just engages the spring. Find the box's kinetic energy at that instant and begin there; as the spring compresses and gains potential energy (spring potential energy), the box loses kinetic energy to the spring but also gains a bit from the continuing drop in elevation.
    What is .3? The initial distance between the box and the spring is 3 meters...
    And ' is usually interpreted to be arc minutes (60ths of a degree). Did you mean to employ the "°" character?
     
  9. Jun 10, 2013 #8
    Sorry i don't really understand. Ive looked at a couple of workings of the problem and the only thing I can't figure out is how to find the height. The distance of the box is initially at the 3m from the spring. But we don't know the length of the spring to the bottom of the ramp?

    Is the length of the spring different from the displacement x when the spring is compressed, (assuming the spring is massless)?
     
  10. Jun 10, 2013 #9
    Okay i think I've got it.
    well the answer is right anyway.
    I think if we assume that the spring is compressed all the way to the bottom of the ramp, transferring all of its kinetic and potential energy to the elastic potential of the spring. PE(final) =0, KE(final)=0

    mgh(initial) + 0.5mv^2(initial) = 0.5kx^2

    where the initial height is (3+x)sin(20)

    (2.5)(9.8)((3+x)sin(20)) + (0.5)(2.5)(0.75^2) = 0.5(500x^2)
    x=0.1314 or =-0.097
    therefore x = 0.1314

    here is a similar solution
    https://www.physicsforums.com/showthread.php?t=443817

    Thanks for your help and patience :)
     
  11. Jun 10, 2013 #10

    gneill

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    Staff: Mentor

    The position of the bottom of the ramp doesn't matter. You aren't given the initial length of the spring either. Neither matters. What is important in these problems involving conservation of energy is the change in height, not the particular height at any given instant. If you know the change in height you can find the change in (gravitational) potential energy.
    The spring will change in length by some Δx. This is independent of its natural length.
     
  12. Jun 10, 2013 #11

    gneill

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    Staff: Mentor

    You don't have to assume anything about the final position with respect to the length of the ramp. The spring will compress until the potential energy it contains equals the available energy from the KE of the box and the energy the box obtains from falling to whatever the end height happens to be. The spring could be kilometers in length, and the answer would be the same.

    The box begins with some initial KE due to its initial velocity, and gains some additional KE as it falls towards the spring. Once it hits the spring, it begins to trade its KE with the spring's PE. Meanwhile, it's still moving downwards as the spring compresses, so there's some additional gravitational PE being transferred (via the block) to the spring.

    Note that you used (3 + x) for the initial height. This acknowledges the fact that the block moves 3m downslope before it encounters the spring. The spring compresses an amount x due to the KE of the block and the energy from the gravitational PE as the block falls further while the spring compresses.
     
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