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Friction with a sliding block

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.8 kg mass (m2) that hangs over the side of the shelf 1.6 m above the ground. The system is released from rest at t = 0 and the 2.8 kg mass strikes the ground at t = 0.81 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 2.8 kg mass now strikes the ground 1.3 seconds later.
    (a) Determine the mass m1.
    kg

    (b) Determine the coefficient of kinetic friction between m1 and the shelf.


    2. Relevant equations

    v^2=v(initial)^2-2ad
    Sum F=ma
    F(kinetic)=u(kinetic)mg

    3. The attempt at a solution
    I applied F=ma to m1 and m2 for x and y

    m1:
    x: T-f(kinetic)=0; T=f(kinetic); T=u(kinetic)m1*g
    y: F(normal)-m1*g=0; F(normal)=m1*g; f(kinetic)=u(kinetic)m1*g

    m2:
    x: T=m2g; m2g=u(kinetic)m1*g

    The next step would be to solve m2g=u(kinetic)m1*g for u(kinetic) or m1 but without one of these I can't find either so I am stuck here.
     
  2. jcsd
  3. Oct 7, 2008 #2

    Doc Al

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    Staff: Mentor

    For some reason, you are treating the system as though the acceleration = 0. Not so. Start by calculating the acceleration for each scenario.
     
  4. Oct 7, 2008 #3
    I did calculate the accelerations actually but I wasn't sure what to do with that. I used v^2=v(initial)^2-2ad and got that the first acceleration is 1.219 and the second is .473
     
  5. Oct 7, 2008 #4

    Doc Al

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    Staff: Mentor

    I'm not sure how you calculated the accelerations using that formula, since you don't have the final speeds. (I see what you did: You used v = d/t for the final speed, but that's not correct. That's the average speed, not the final speed. The final speed is twice that.)

    Recalculate the accelerations, either by correcting your value for the speed or by using a different formula entirely (one that uses distance and time, which is what you are given).

    You need the accelerations for use with Newton's 2nd law.
     
  6. Oct 7, 2008 #5
    Okay thank you! i found the accelerations using v=d/t and then a=change in v/t. But now I am trying to plug it into a=F(net)/m and don't see how to find this without knowing either u(kinetic) or mass. The sum of the forces would be F(normal)+F(kinetic)+tension

    F(normal)=mg
    F(kinetic)=u(kinetic)*F(normal)
    Tension=?

    I'm not sure how to find tension. I think once I do I can plug the sum of the forces into a=F(net)/m and then the masses will cancel and I can get u(kinetic)? is this correct?
     
  7. Oct 8, 2008 #6

    Doc Al

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    Staff: Mentor

    Again, since the system is accelerating, v = d/t gives you the average speed, not the final.
    The mass and μ are things you are asked to find. You're going to solve for them.
    Make sure you sum the forces in each direction separately.

    You need to apply ΣF = ma to both masses. You'll get two equations for each scenario, which you'll combine to eliminate tension.
     
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