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Kinematics ramp and crate problem

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data

    You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1480 N will move with speed 2.1 m/s at the top of a ramp that slopes downward at an angle 23.0 degrees. The ramp will exert a 578 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.
    Calculate the maximum force constant of the spring k_max that can be used in order to meet the design criteria.

    2. Relevant equations

    hinital= (sin23=h/8) = 3.126
    hfinal= 0
    (spring)xinitial= 0
    (spring)xfinal= ???
    vfinal= 0
    vinitial= 2.1

    Wfriction = Ffriction * Distance

    Change in gravitational potential energy = mghfinal-mghinitial

    Change in elastic potential energy = (1/2)kxfinal2 - (1/2)kxinitial2

    Change in Kinetic Energy = (1/2)mvfinal2-(1/2)mvinitial

    3. The attempt at a solution

    Since the final velocity, final height, and initial x for the spring are all equal to zero I got the equation..

    Wfriction + mghinital + (1/2)mvinital2 = (1/2)kxfinal2

    And then I solved for k..

    k= (2(Wfriction+mghinital+(1/2)mvinitial2))/x2


    I don't know if what I did it's right, but if it is.. I can't solve for k since I don't know x, the distance that the spring compressed, and I don't know how to find it =[[
    Help pleaseee
     
  2. jcsd
  3. Oct 23, 2009 #2
    this is perfect, although bear in mind that the work against friction will be negative (because it's energy that it has lost).

    what I would do from here is put in the values that you know in order to get a nice equation relating k & x.

    next you can set up a simultaneous equation, because you also know hooke's law don't you? that the force is proportional to the spring constant multiplied by the extension

    F = kx, you know what K is (bear in mind F is a vector - if it's still on the ramp the only bit the spring is supporting is the horizontal component of the force)

    do you see why this works?
     
  4. Oct 23, 2009 #3
    I think so.. but what is the force of the spring in hooke's law?
     
  5. Oct 24, 2009 #4
    Hooke's law is as follows

    the force impressed on a spring is equal to the extension of the spring caused by that force multiplied by the spring constant.

    so F = kx

    however like I said bear in mind that the spring is on the angle and is only supporting the box's horizontal motion (the floor is supporting the box vertically) so perhaps it will look something like

    mg cos(theta) = kx

    have you set up the simultaneous equation yet? you had the right formula you just needed to put in the numbers.
     
  6. Oct 24, 2009 #5
    Okay, so I set..

    mgcos(theta)=k*x

    my equation for K was..
    (2(Wfriction+mghinital+(1/2)mvinitial2))/x2

    And i got x= .49 m
    which I then substituted into my equation for k, but I got the wrong answer
    is my equation for k wrong?
     
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