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Coefficient of kinetic friction?

  1. Oct 8, 2004 #1
    Okay, I've never done this before and I sure hope there is somebody out there who can help me. I've tried answering this question EVERY way possible and cannot come up with the right answer.

    In a circus performance, a monkey is strapped to a sled nd both are given an initial speed of 4 m/s up a 20º inclinedj track. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?
     
  2. jcsd
  3. Oct 9, 2004 #2

    rcgldr

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    The amount of mass doesn't matter, so your just dealing with deceleration due to gravity (constant) and friction (also contant in this case).
     
    Last edited: Oct 9, 2004
  4. Oct 9, 2004 #3

    Pyrrhus

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  5. Oct 9, 2004 #4

    rcgldr

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    Well that solution is correct, but it's more complicated than needed.

    Deceleration from constant fiction is equal to the coefficient of friction times the rate of accleration of gravity. The friction is relative to the component of gravity normal (perpendicular) to the track

    or .2 x cos(20).

    So the rate of frictional decleration is

    .2 x cos(20) x 9.8 m / s^2.

    The component of gravitation deceleration is relative to the component parallel to the track:

    sin(20) x 9.8 m / s^2.

    Total deceleration rate is the sum of these 2 rates

    .2 x cos(20) x 9.8 m / s^2. + sin(20) x 9.8 m / s^2 = 5.3 m / s^2.

    Since the initial speed is 4 m/s it takes (4 / 5.3) ~= .75 seconds to stop.

    Since it's linear deceleration, the average speed is 1/2 the inital speed, in this case 2 m/s, so the total distance traveled is 2 x .75 ~= 1.5m.
     
    Last edited: Oct 9, 2004
  6. Oct 9, 2004 #5

    rcgldr

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    Ok, math wizards, what if the track were curved with radius R? Assume the track starts off horizontally, and R is large enough that the track doesn't go beyond veritcal before the sled stops.
     
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