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Work and Kinetic Energy equation

  1. Oct 14, 2004 #1
    I have this problem to do and am confused on it. Here is the problem statement, word for word (and the diagram attached):

    "The small slider of mass m is released from rest while in position A and then slides alond the vertical-plane track. The track is smooth from A to D and rough (coefficient of kinetic friction, mu_k) from point D on. Dertermine

    a) the normal force N_B exertered by the track on the slider just after it passes point B

    b) the normal force B_C exerted by the track on the slider as it passes the bottom point C

    C) the distance s traveled along the incline past point D before the slider stops.

    I think I can do C and probably B, but first I need to know how to do part a. I have no idea where to start. I would think that the normal would be zero since it is vertical..but of course that is where it starts to change direction...

    I must use this equation:
    sum of the work = change in kinetic energy

    Please help me someone

    (view attachment for pic)
    Sorry for the poor quality.....I had to make it way smaller since the picture can only be so big to upload on this site....the angle is 30 degrees, the radius is R, and the length on the left is 2R

    Attached Files:

    Last edited: Oct 14, 2004
  2. jcsd
  3. Oct 14, 2004 #2
    Right at point B, you are correct, the normal force should be zero. It is only after this point when it starts to increase.

    For the normal force at point C you will need to take into account both the acceleration of gravity as well as centripetal acceleration.
  4. Oct 14, 2004 #3
    but it wants N_B "right past point B"

    the answer for part a is 4mg
  5. Oct 14, 2004 #4
    Ah, I see what they are getting at. Yes, you are correct, "right past B" is to be considered the point when you have to take into account centripetal acceleration since it is now on the curved portion of the track.

    So, if you get an expression for velocity at this point, and then work it into centripetal acceleration, you end up with the answer they give.

    Sorry about the mistake.
  6. Oct 14, 2004 #5
    oh, ok....I was able to figure it out, thanks for the help!
  7. Oct 14, 2004 #6
    Here is another one I am having some trouble with. Can someone point me in the right direction?

    "The spring mounted .8-kg collar A oscillates along the horizonal rod, which is rotating at the constant angular rate theta-dot = 6 rad/sec. At a certain instant, r is increasing at a rate of .8 m/sec. If the coefficient of kinetic friction between the collar and the rod is 0.40, calculate the friction force F exerted by the rod on the collar"

    See the attachment (sorry for the poor quality :(...I will do better next time when I get a better program to work with!

    I need to find out what r is at this instant and what r-double dot is...does anyone know what I can do? I have tried tons of different ways, but I keep messing up.....this is just a simple problem of using F=ma...applied i nthe r direction...thus:

    F (friction) = m(r-double dot - r * theta-dot^2)

    I know theta dot is 6, and I know m. But I need to find r-double dot and r

    Any ideas?

    Attached Files:

    Last edited: Oct 15, 2004
  8. Oct 14, 2004 #7
    anyone? I need to figure this problem out. Also any ideas for part C on the first problem would be helpful.
  9. Oct 15, 2004 #8
    ok, I got part C from above.....

    I just need help on this last one, please help me someone
  10. Oct 15, 2004 #9
    ok, I think I finally figured it out, so nevermind!
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