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A physics problem.

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A 31 kg skier skis directly down a frictionless slope angled at 13° to the horizontal. Choose the positive direction of the x axis to be downhill along the slope. A wind force with component Fx acts on the skier. What is Fx if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of 1.2 m/s2, and (c) increasing at a rate of 2.4 m/s2.

    2. Relevant equations

    Newton's second law.

    3. The attempt at a solution

    The velocity is constant hence acceleration tends to 0. The forces acting on the skier are the normal force, weight, and the wind force. In the first part, the net force is equal to 0, therefore the force (Fx) is -68.34 N.

    I attempted to do the second and third part. I assumed that the net force will be equal to the mass times the acceleration, which is 1.2 and 2.4 respective of the second and third parts. I found two values for Fx but the site and my instructor said they were wrong. I tried solving it many times and it gave the same answer.
  2. jcsd
  3. Oct 9, 2011 #2
    Use F(net) = ma down the plane.
    If you show your working people can help you.
  4. Oct 9, 2011 #3
    I did use it and found out that in part b, Fx = 31 N and in part c, Fx = -6.2 N.

    The net force is equal to ma, then: (@ is theta)

    -Fx + mgsin@ = ma

    with a = 1.2 m/s2, Fx = 30 N
    with a = 2.4 m/s2, Fx = -6.2 N

    Is that true? And is the projection of mg along the x component mgsin@ or its negative? I took the positive direction down the inclined plane.
  5. Oct 9, 2011 #4
    Fx = 31N is OK
    The projection of mg along the x component is not negative since mg is vertically downwards and you have chosen positive x dir to be down along the plane.
    As regards Fx = -6.2N, try to find the max acceleration with no wind at all.
  6. Oct 9, 2011 #5
    The maximum acceleration, according to my quick calculations, is roughly 2.2 m/s2, which is greater than 2.4 m/s2. There's something here, and I'm guessing: will Fx be zero then? Please bare with me.

    The whole thing is just vague to me considering I haven't been perpetuating the subject, mainly motion.
  7. Oct 9, 2011 #6
    Never mind. Figured it out.
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