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Springs: Inclined ramp, object moving down towards spring

  1. Jun 27, 2007 #1
    1. The problem statement, all variables and given/known data

    You are designing a delivery ramp for crates containing exercise equipment. The crates weighing 1490 N will move at a speed of 2.10 m/s at the top of a ramp that slopes downward at an angle 25.0°. The ramp exerts a kinetic friction force of 540 N on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 7.70 m along the ramp. Once stopped, a crate must not rebound back up the ramp.

    Calculate the force constant of the spring that will be needed in order to meet the design criteria.

    2. Relevant equations

    I have been using:

    [tex]K_{1} + U_{grav1} + U_{elastic1} - W_{friction} = K_{2} + U_{grav2} + U_{elastic2}[/tex]

    3. The attempt at a solution

    [tex]K_{1}= (1/{2})(1490/{9.8})(2.1^{2})=335.25 J[/tex]
    [tex]U_{grav1} = U_{elastic1} = 0[/tex] (because at the top, I have y=0)
    [tex]W_{friction} = (540)(7.7) = 4158 J[/tex] (force * distance)

    [tex]K_{2} = 0[/tex]
    [tex]U_{grav2} = (1490)(-7.7)(cos 25) = -10398.1 J[/tex]
    [tex]U_{elastic2} = (1/2)(k)(7.7^{2})[/tex]

    So, my first thought is, I have no idea how long this spring is, so I have no clue how much it is being compressed. Would I be correct to say that x=(7.7)^2 in the U_elastic equation?

    Using this and solving for the k in the 2nd elastic equation, I am not getting a correct answer. I am getting 221.803.

    This is what I input into my calculator, so you can see how I arranged the equation to come up with K:

    Please point out my error, thanks a lot.
  2. jcsd
  3. Jun 27, 2007 #2


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    I see one error here:

    [tex]U_{grav2}=mgh=(1490)(7.7\sin(25))[/tex], not cos(25).

    See if this helps. With regard to the length of the spring, it seems weird that they don't give you any information about it. Are you sure you don't know anything about it? Try to use your current assumption and see what happens now with the new trig function...
  4. Jun 27, 2007 #3
    Thanks. What I posted was the entire text of the problem as it was given to me. There isn't any additional information.

    So we have (1490)(-7.7)(sin 25) = -4848.7.

    If I substitute your correction into the formula,
    (335.25-4158+4848.7)*2/(7.7^2), I get 34.6079, which is something I actually tried earlier and that wasn't correct either.

    (I assume you meant (1490)(-7.7)sin(25), otherwise when I move it into the left hand side, I get a negative number result for k, -292.51).
  5. Jun 27, 2007 #4


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    Ok. So, the information that neither of us has used yet is that we want the crate to stay put after it is stopped. This means that the net force on it at this point must be zero. Can you set up an equation describing this? If you can you'll be left with two equations and two unknowns. Can you take it from here?
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