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Homework Help: Just needing a little help with physics and springs

  1. Apr 1, 2004 #1
    ok, ive been at this for quite some time. its pretty difficult for me and need some help fast.

    The question is
    A 2kg box lies at rest on an incline of 30 degrees with a friction of .2 . The box slides down, and compresses a spring of k=10n/m. What is amount of forces that shoots up the box and at what speed? How far does the box go?

    picture of the problem is at
    http://panda.rightclicked.net/untitled.bmp [Broken]

    Just need to know how exactly i would go about this, because I am not sure what equation to use.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Apr 1, 2004 #2
    At first look it seems you will also need the height of the ramp, or the distance that the box travels before hitting the spring, in order to solve this question. Assuming you have this information, you can solve this problem using energies and work.

    [tex]W_{friction} = \Delta E_m = \Delta E_k + \Delta E_{p_{gr}} + \Delta E_{p_{ele}}[/tex]
    The work of friction is equal to the change of mechanical energy in the system, which is equal to the sum of the changes of kinetic, gravitational potential and elastic potential energies.
  4. Apr 1, 2004 #3
    sry missed a couple of info.

    The spring is at normal rest and length = 5cm, and then compressed 2.5cm.
    I need to find the forces that the spring releases the box at, and at what velocity it is, then how far up the ramp the box goes.
  5. Apr 1, 2004 #4
    Ok, so you know how much it compresses (let's call that l) and you can thus find the force it exerts on the box (kl). You can also find the elastic potential energy of the spring, 1/2kx^2. You know that the kinetic energy of the box is 0 both up and down, since it is at rest at those points. Lastly you need to find the normal force on the box so you can find the magnitude of the friction force. Once you have all of that you plug it into the equation I posted above to get:

    [tex]-fx = (0 - 0) + (mgh - 0) + (0 - \frac{1}{2}kl^2})[/tex]

    Where [tex]h = x\sin 30[/tex]. Solve the equation for x and you're done...
  6. Apr 1, 2004 #5
    what are the equations i can use with the angle??

    for Kinetic energy would it be, (1/2)mv^2 cos (theta) ?? or would it be sin?
    for Potential Energy : 1/2 k x^2 ?? no angel into account?

    and for the force do i take in the angel to account?? like, mgcos(theta), so would it be, kl cos(theta)?
  7. Apr 1, 2004 #6
    ok heres the deal...

    the force the spring creates on the box can be done by using f=kx f=10n/m (.025m) f=.25N (Newtons)

    the normal force must be found using mgcos(theta) (2kg)(9.8m/s^2)(cos30)

    N=9.8(2)(.866) N=16.97


    apparently the box wouldnt move... trick question?
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