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Homework Help: Work-energy problems

  1. May 20, 2005 #1
    Hello. I'm having a lot of trouble with two problems from a homework assignment. I've spent a good amount of time on both problems but just can't seem to get an answer! A step-by-step or suggestions would be most helpful. Thanks in advance!

    1. A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill. The crest of the second hill is circular, with a radius of r = 25 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

    2. An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 12.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.60 m below the edge. How fast is she going just before she lands?

    Maybe I just can't understand the physics of skiing! :uhh:
  2. jcsd
  3. May 20, 2005 #2


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    1. First you need to find what velocity the skier would need to have to just lose contact with the hill. Gravity is the centripetal force, and must be greater than mv^2/r to keep the skier on the ground. Once you know this velocity, use conservation of energy to find the height that the skier would have to descend to build up this much kinetic energy.

    2. This one is a little trickier. First you need to get the velocity the skier has built up when she clears the edge of the cliff. The easiest way to do that would be to use conservation of energy. The kinetic energy gained is equal to the potential energy lost minus the work done by friction, which is equal to Fd, the magnitude of the force of friction times the distance over which it acts. Once you have this velocity, you can do the second part. Ignore the horizontal motion for now. Figure out how much speed would be built up by someone (with an initial velocity equal to the vertical component you just found), on a 3.6m fall. Add this to the vertical component, and finally, use pythagoreans theorem to get the magnitude of the final velocity, using this new vertical component and the horizontal component from before.
    Last edited: May 20, 2005
  4. May 20, 2005 #3
    Sorry, I should probably emphasize. The particular problem I'm having with the first question lies more within the velocities of the conservation of mechanical energy equation, specifically the final velocity (since we KNOW the initial is zero as the skier starts from rest). I don't see how I should go about applying the information of the skier losing contact with the snow...

    As for the second, I'm convinced that I need to know the mass of the skier before I can do anything with this problem...or do the masses cancel? I've tried a few ways and wasn't able to come up with anything.

  5. May 20, 2005 #4

    Thanks for the reply. The m (mass, I'm assuming) of the skier was not given in the problem :devil: how should I go about determing this mass to find the velocity of the skier?
  6. May 20, 2005 #5


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    Don't worry about the skier leaving the ground. You're trying to find the speed where she just leaves it, so you can safely pretend she is always on the ground (I don't know if you know calculus, but you would be considering the limit as she approaches the speed where she would actually leave the ground).

    For the second, yes, the masses cancel. Friction is proportional to the gravitational force, which is proportional to mass, so the total force will have an m out front, and since F=ma, m will cancel and you can go on to compute a, v, and anything else. (you wouldn't actually use F=ma (conservation of energy is much easier); this is just to show why it works).
  7. May 20, 2005 #6
    Ok First Of All:

    When the skier loses contact on the second hill,the N(normal reaction due to hill) vanishes.Consider the forces acting on skier at the top of the hill.

    1]Mg acts downwards
    2]N upwards

    [itex]Mg=m \frac{v^2}{R}[/itex]

    There is no N in the above equation because it vanishes when the skier leaves the contact
    [itex]V_f = \sqrt gR[/itex]

    Now apply Energy conservation at 2 pts at the top of two hills:

    mgH=mg(25) + \frac{1}{2} mV_f^2

    Calculate H from above

    Last edited: May 20, 2005
  8. May 20, 2005 #7
    Allllright...thank BOTH of you for the help on figuring out the first problem. Got the answer of h=12.5, which my WebAssign agrees to be correct. Now I'll play with the second problem!
  9. May 20, 2005 #8
    Problem 2, we're close. Very close. The horizontal motion part I understand no problem. Still having trouble with the vertical motion down the hill, however. I understand the setup for the equation Status described above, but how should I go about determining the Fd frictional force that should be subtracted from the grav. potential energy?
  10. May 20, 2005 #9


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    F, the force of friction, is equal to μN, where μ is the coefficient of friction (0.2), and N is the normal force, equal in magnitude the the component of gravity perpendicular to the slope.
  11. May 20, 2005 #10


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    Hi nutster. I'm curious about your answer to the first question. Did the problem say that the top of the second hill was at the same height as the bottom of the first hill?

    That's what your answer suggests. Just wondering. When reading the question I was unsure as to the height of the second hill. Was there a picture given?
  12. May 20, 2005 #11
    OK, I thought I had it but I'm still having some issues with it, and sadly the assignment is due very soon. Here's the equation:


    I find Fd to be equal to 22.02...is this correct? The only other thing I'm having trouble finding is h, the height...I think it's 5.24m.
  13. May 20, 2005 #12


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    Your equation should be:


    Then plug in F=uN = u (mg cos(25))

    So you get

    .5m(Vf)^2 = (mgh) - u (mg cos(25)) d

    The m's cancel

    .5(Vf)^2 = (gh) - u (g cos(25)) d

    h is the total height dropped all the way from the top... so it's d sin (25) + 3.6 = 5.24 + 3.6 = 8.84m

    So now you can solve for Vf.
  14. May 20, 2005 #13
    OK, so I find Vf to be equal to 11.367...I actually had this much figured out. I think my problem is I'm not sure what to do at this point (i.e. how to integrate it into the rest of the problem). How do I use this in conjunction with the fall at the end?
  15. May 20, 2005 #14


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    That's the final answer isn't it? I'm not sure what else there is to do... You've solved for the velocity right before she lands.
  16. May 20, 2005 #15
    Right you are! See, told you I didn't know how to integrate it into the problem! Thanks for the help, all of you. It's much appreciated. Might see you again soon!
  17. May 20, 2005 #16


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    woops, i gave some bad advice. conservation of energy can solve the whole problem.
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