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Kinetic Friction, did I do this properly?

  1. Dec 10, 2007 #1
    I was hoping somebody could check over my work and see if I have the correct answer for this problem, as my text hasn't been the greatest in explaining this.

    QUESTION
    A sight seen on many bunny hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate

    a) the time taken for the skier to come to a stop
    b) the distance travelled down the hill

    MY ANSWER
    Fa = Fgh - Ff
    = Fg sin 5(degrees) - ukFn
    = mg sin 5(degrees) - uk mg cos 5(degrees)
    = (25kg)(9.8 m/s^2)(0.08716) - (0.20)(25kg)(9.8 m/s^2)(0.9962)
    = -27.5N

    F = ma
    a = -27.5N / 25kg
    a = -1.1m/s^2

    delta t = (vf - vi) / a
    = (0 - 3.5m/s) / -1.1m/s^2
    = 3.2s

    a) Therefore it takes the skier 3.2 seconds to come to a stop.

    delta d = vi(t)+0.5(a)(t)^2
    = (3.5 m/s)(3.2s) + 0.5(-1.1m/s^2)(3.2s)^2
    = 11.2m - 5.632m
    = 5.568m
    = 5.6m

    b) Therefore the skier travelled 5.6 meters down the hill.

    Thanks so much for your time, I really appreciate it!
     
  2. jcsd
  3. Dec 10, 2007 #2

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    Quite correct. However, note that there was no need to multiply the mass to find Fa, since in the next step you divided by m to find a. Next time you may save time by using only symbols until the last step.
     
  4. Dec 10, 2007 #3
    The whole procedure is perfect.

    But...
    The person does apply some force as he gives some initial velocity to himself.

    So you have to consider that also.
     
  5. Dec 10, 2007 #4

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    It has been taken into consideration in the value of the initial velocity.
     
  6. Dec 10, 2007 #5
    Great thanks so much for the help! I was hoping I was on the right track, unfortunately the text I must use is not ideal :(
     
  7. Dec 10, 2007 #6
    No

    That is taken for the kinematics equation.

    We do have to consider the force which the person applies.

    His acc is not merely due to mgsin but also due to the force which the person applies.
     
  8. Dec 10, 2007 #7
    If you can afford then go for Halliday Resnick.
    It is the best book for physics.
     
  9. Dec 10, 2007 #8

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    His initial speed is due to an impulsive force, with which he has pushed off and which had given him an initial momentum. After that, this impulsive force has ceased to exist, and the dynamics of motion is solely governed by m, g, uk, angle of the slope and initial velocity.
     
  10. Dec 10, 2007 #9
    Ok then lets talk about a horizontal skiing surface. In case of an horizontal surface there is no mgsin so why does in that case the person is able to move though it would be a short distance?
     
  11. Dec 10, 2007 #10

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    Because of the initial momentum imparted to the person due to an impulsive force created by whatever means. He may push at the ground by means of the ski poles, or somebody may push him. If he does not have any other means of propulsion, ultimately he'll come to rest due to friction.
     
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