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Kinetic Energy and Work-Energy Theorem

  1. Oct 5, 2004 #1
    Hello,

    I could use a little guidance on this problem.

    An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25 degrees with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 13.1m before comming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.30m below the edge. How fast is she going just before she lands?

    So here is my reasoning and what I have done thus far. I drew my free body diagram which really doesn't really do anything for me, but I know it is supposed to. Anyway, I also drew a little picture of what the hill and skier might look like.

    I decided that to find the initial height I would use 13.1m(sin 25) to get the inital height. I know that the final height is 3.30m. I also know that initial velocity is 0 since she is starting from rest and I'm left to find the final velocity. So here's my little chart.

    Ho = 5.54
    Hf = 3.30
    Vo = 0
    Vf = ?

    So forgetting about the friction for a moment I found the velocity the skier would be travelling at the landing point using this equation: Vf = square root of Vo^2 + 2g(Ho - Hf)

    But I know that isn't the answer because I didn't take the friction into consideration. Thus my question. How do I fit that in there? This is the part that is stumping me. Perhaps I should have used a different equation to include the friction but I don't have the slightest clue what that could be.

    Any help provided would be greatly appreciated.
     
  2. jcsd
  3. Oct 5, 2004 #2
    Try this. Use simple kinematics/Newton's laws formulas. What is the force (net, of course) acting on the skier while she is still on the inclined plane? Can you find her final speed when she reaches the cliff using this? Now it's a projectiles problem.
     
  4. Oct 5, 2004 #3
    Well that could work, but I'm supposed to learn this using the Work-Energy Theorem and the Kinetic Energy equations.
     
  5. Oct 5, 2004 #4

    Gokul43201

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    So what does the work-energy theorem say ? This is a direct application of the equation; you just have to plug in the numbers to get the answer.

    And you have the wrong value for Ho. Should it not be (5.54 + 3.30) m ?
     
  6. Oct 5, 2004 #5
    Mmm, even better. You have the right approach, but factor in the work done by friction. Keep the variable for mass in the equations (as you are not given the skier's mass), and see what happens...

    You should be able to finish the problem once you get her velocity at cliff's edge.

    EDIT: I was writing this while Gokul posted...
     
  7. Oct 5, 2004 #6
    Well I would just plug in the numbers but the mass is unknown. Without the mass I can't find the net force.

    The work energy theorem is: W = 1/2mVf^2 - 1/2mV0^2

    And with that equation the initial height and final height are not included. I can't understand why I'm having such a hard time with this problem. I just did my other problem without a hitch but this one has friction involved which is throwing me off.

    Oh and yes, it would make more sense to add the final height to the height I found to give me the correct amount for the initial height.
     
  8. Oct 5, 2004 #7
    Think about the energy transfers going on here. You begin with only gravitational potential energy. During the trip to the cliff's edge, what happens do different portions of the initial potential energy of the skier?
     
  9. Oct 5, 2004 #8

    Gokul43201

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    What Sirus said, and in addition, I'd like to say that your equation for the work energy theorem is incorrect.

    Here's the correct one :

    Total initial energy - work done = total final energy.

    Keep in mind that total energy = PE + KE.

    Now proceed. (just call the mass m and write down all the terms in the equation. But the final height is not 3.3m, it should be 0m, if the initial height is 8.8 m. However, if your initial height is 5.5 m, the final height should be -3.3 m. Always draw a picture and dimension it correctly.)
     
    Last edited: Oct 5, 2004
  10. Oct 5, 2004 #9
    Well that's how my book printed that equation. KE = 1/2mv^2 so my equation was correct. Just the extended version.

    And like I said, I did draw a picture, but honestly it does nothing for me. I know it should, but it doesn't. I will try what Sirus said, but I still don't know how to factor in the friction as he mentioned. And if I'm leaving m in the equation how does that let me solve for the final velocity? I will then have 2 unknown variables.
     
  11. Oct 5, 2004 #10

    Gokul43201

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    Your equation would be correct if you also included the initial and final PEs.

    Do you know the initial KE ? What about the initial PE ? Write down the expressions for these.

    Similarly write down expressions for the final values.

    Now what is the work done = ? Write this expression down too. Now simply plug all these into the work-energy theorem and you are done.

    Show the working here, and I'll help you along...
     
  12. Oct 5, 2004 #11
    I still don't follow but here goes. This makes no sense to me since mass is unknown. But here goes.

    KE = 1/2mv^2
    KE = 1/2m(0m/s)
    KE = 0

    PE = mgh
    PE = m(9.80m/s)(8.8m)
    PE = 86.24m

    Work done I'm guessing is:
    W = mg(Ho - Hf)
    W = ( m(9.80)(5.5 - (-3.30)) )
    W = 9.80m(8.8)
    W = 86.24m

    Ok, I'm a bit stuck from here. Though, and I still don't see where friction comes into this.
     
  13. Oct 5, 2004 #12

    Pyrrhus

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    [tex] \Delta E = W_{f} [/tex]

    Look up that equation, should be on your book, or at least it should talk about NonConservative forces and mechanical energy.
     
  14. Oct 5, 2004 #13
    Thanks Gokul for trying to help me. I really do appreciate it. Unfortunately, I'm off to bed and won't make it back on before my class tomorrow. I'll either not answer it or hopefully I can find someone in my class who can help me work through it sometime before it's due.

    Thanks again, I know you will be seeing me around here soon. LOL
     
  15. Oct 5, 2004 #14

    Gokul43201

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    Okay, now write down the values/expressions for the final KE and PE.

    Okay, the "work done" IS where friction comes in.

    The work done is the force of friction multiplied by the distance over which it acts. What friction does is, it converts mechanical energy into heat. This is just a way of accounting for the loss of energy.

    So calculate the work done (by friction) using this.

    PS : Too late, I guess.
     
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