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Spring calculation problems

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data

    Skjermbilde 2015-10-14 kl. 19.41.21.png
    A man is jumping up and down on a pogo stick. For the purpose of this

    problem, we will think of such a stick as being a (massless) spring with a guy

    + stick on top. We will assume that no friction is involved, and so in principle

    the guy could keep jumping forever. At his highest point, the bottom of the

    (unstretched/unsqueezed) spring is at a height of 40.0 cm from the ground.

    The spring constant of the pogo stick is taken to be k = 50000 N/m. The

    mass of the guy is 70 kg and of the pogo stick 5 kg. Ignore details about him

    being able to go higher and higher by bending his knees and stuff; he just

    stands on the thing.

    a) Explain, using a energy bar chart how the different types of energy increase

    and decrease at different stages of the jumping motion. What is the total

    energy of the system? (For the purpose for the gravitational potential energy,

    this is taken to be relative to the lowest point of the jumping motion, when

    the spring is on the ground, and the spring is squeezed; think carefully about

    this.)

    b) Compute the value of the different energy components at the following

    points: At the highest point of the jumping motion (A); At the lowest point

    of the jumping motion (B); at the point where the spring of the pogo stick

    is at its equilibrium point (C).


    3. The attempt at a solution
    a) When the spring is on the ground and squeezed, the spring has 100% Potential energy
    When the spring is at equilibrium; the spring has 100% Kinetic energy
    When the spring is 40 cm above the ground, the spring has 100% Gravitational potential force.
    On the way back down, when the spring touches the ground, the spring has 50% Kinetic and 50% Potential energy.

    b) (A)=294J
    (B) -xmax=-0.0015m
    U=0.056J
    (C) x=0
    U=0


    Are my solutions correct?

    All answers are appreciated!1
     
  2. jcsd
  3. Oct 14, 2015 #2

    haruspex

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    Think again about how the energy is distributed when the spring is at equilibrium.
    (In all of those answers, you discuss the energy of the spring. You mean the energy of the man+stick+spring system.)
    Are you sure it's 50/50 between KE and PE when the spring touches the ground? How did you arrive at that? Please post your working.
    (It may be right, I have not checked.)

    For part b, you are supposed to give the energies for each energy type at each of the points.
    For the PE, read very carefully what the question says about where the zero PE height is taken to be.
     
  4. Oct 15, 2015 #3
    How did you find (B) and (C) and also spring at equilibrium point, isnt that when the spring neither is stretched or squeesed (x=0), but how did you get -0.0015m in B?
     
  5. Oct 15, 2015 #4
    I did it this way:
    -xmax=mg/k=-0.0015m
     
  6. Oct 16, 2015 #5
    I've got total energy (since ground is said to be height 0.40m) 385 J using mgh+mgx where X=0.124
     
    Last edited: Oct 16, 2015
  7. Oct 16, 2015 #6

    haruspex

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    Mg/k gives the spring compression at equilibrium. The lowest point of the bounce is not equilibrium.
    Suppose the spring has relaxed length L and its maximum compression during the bounce is x. At the highest point of the bounce, the bottom of the spring is height H=.4m from the ground. How high is the top of the spring from the ground then? At the lowest point of the bounce, how high is the top of the spring from the ground?
     
  8. Oct 16, 2015 #7
    So energy on the top will be (also the total energy, since energy is conserved) Ep=mg(h+x), where x is the distance between the lowest point of the spring and the heighest point of the sprinng?
     
  9. Oct 16, 2015 #8

    haruspex

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    I agree with your answer in post #5. I also agree with your post 7, except for how you defined x. If you define it the way I did in post 6 then the energy is mg(H+x).
     
  10. Oct 16, 2015 #9
    Thanks for the help. I agree i could define it better.
     
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