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What is the distance moved up the plane (friction problem)

  1. Jun 3, 2009 #1
    A particle P of mass 3 kg is projected up the line of greatest slope of a plane inclined at an angle of 50° to the horizontal. The coefficient of friction between P and the plane is 0.5. The intiial speed of P is 9 m/s.

    ii) Find the frictional force acting on P

    ii) What distance moved up the plane by P until its velocity becomes zero.


    2. Relevant equations

    f = μ x N where N is the normal

    N = m x g x cos(50)


    3. The attempt at a solution

    i)

    N = m x g x cos(50) = 19.28 N

    f = 0.5 x 19.28 = 9.64 N


    ii) Here I do not know what to do. I thought I could take -5 m/s^2 as a deceleration by using :

    f = μ x N

    m x a = 0.5 x m x g

    a = 5 m/s^2

    and then using SUVAT equations but this is wrong since I do not get the answer shown in my book..

    Please help!

    Thank you!
     
  2. jcsd
  3. Jun 3, 2009 #2

    tiny-tim

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    Last edited by a moderator: Apr 24, 2017
  4. Jun 3, 2009 #3
    I did think of that..but shouldn't friction still be taken into account? Because I did the following :

    1/2 x m x v^2 = mgh

    h = v^2 (2g) = 4.05 where v = 9 m/s and g = 10

    and then in order to find the hypotenuse i used

    Hypotenuse = 4.05 / sin(50)

    But this gives me 5.28 m and the answer is 3.72 m
     
  5. Jun 3, 2009 #4

    tiny-tim

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    Last edited by a moderator: Apr 24, 2017
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