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Spring force problem

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data
    The diagram shows a potato launcher before the spring is compressed and used to launch a potato. A force of 65 N is required to compress the spring 5 cm. The spring will be compressed 10 cm before the potato is inserted. The inside of the tube and the potato skin have a kinetic coefficient of 0.15 . The potato has a mass of 180g.

    What is the maximum range of the potato if it launched at 50 degrees with respect to the horizontal? You may ignore air resistance.


    2. Relevant equations

    Eo + Ef = W(non conservative)
    (PEo + KEo) + (PEf + KEf) = W(non conservative)
    F = ma
    Vf = Vo + at
    Vf^2 = Vo^2 + 2aΔx

    3. The attempt at a solution

    So what I did is drew an free body diagram of the potato before launch, the distance the potato will be traveling along the launcher would be 1m if the spring was compressed 10cm, so what that said I applied Work with non conservative force equation [(PEo + KEo) + (PEf + KEf) = W(non conservative)] , correct me if I am wrong but I replaced the potential energy with the Spring force equation which is Fs = -kx with k being equal to 65N, I then solved for Vf which im assuming will get me the range of how far the object will go maybe by using a kinetic equation, but I don't see how I can do that. Am I looking for the final velocity which will get me the distance or is there something else I need to find.

    thanks for any/all your input!

  2. jcsd
  3. Apr 1, 2013 #2
    I think your idea of thinking about the energy is the way to go. But I think you need to add in the potential energy stored in the spring when it's compressed and not the spring force. -kx + 1/2mv^2 doesn't make much sense to me because you're adding Newtons to Joules.

    For the first part of the launch (when the potato is being pushed by the spring) you'll want something like...

    work done by the spring - gain in GPE of the potato - work done against friction = final KE of potato

    I'd start off by using this equation to find the speed of the potato when it loses contact with the spring.
  4. Apr 1, 2013 #3


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    If you have correctly calculated the exit velocity (I didn't check that) then you can resolve it into horizontal and vertical components. Then apply the standard equations of motion in the vertical to calculate the flight time.
  5. Apr 1, 2013 #4
    Thanks for your input! This is what I have so far.


    And to solve for t do I use the equation

    Vf = vo + at with a being gravity then just use that time to find the range?
  6. Apr 1, 2013 #5
    Hey did you happen to get 6.3740m?
    Last edited: Apr 1, 2013
  7. Apr 1, 2013 #6
    How did you get 65N for your value of k? The units for k are N/m, correct?
  8. Apr 1, 2013 #7
    In the op it says the force required to compress the spring was 65N and since work is in joules I assumed that the work done by spring = spring force x distance. Cause if k was already in joules wouldn't that mean it would be joules x m which is an incorrect form of unit right?
  9. Apr 1, 2013 #8
    Hey do you happen to have the answer to this? I would like to check my work. :)
  10. Apr 2, 2013 #9


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    That only works for constant force. More generally, E = ∫F.dx. For springs, F = kx, where k is the spring constant (and F is the force exerted on the spring, so has the same sign as x). That yields E = kx2/2. Use the force and distance given to find k, then use that and the distance to find E.
  11. Apr 2, 2013 #10
    So I found k to be 1300kg/s^2 by doing 65N = k(0.05m), and I plugged that into your equation which I'm assuming E means the work done by the spring and I have the same answer.. is there something I did wrong?
  12. Apr 2, 2013 #11


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    It just happens to give the same answer because the distance it's compressed for the launch happens to be exactly double the sample distance in the info provided. If you'd been told that compressing it distance x took a force F, but in the launch it will be compressed y, the constant would be F/x, and the energy for the launch would be (F/x)y2/2. If y=2x, that gives 2Fx, so happens to be the same as Fy.
    After that, your work so far looks ok.
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