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This problem is driving me crazy. It involves an incline, two masses and a rope.

  1. Feb 14, 2009 #1
    This has been driving me crazy because I thought I knew what I was doing, but the book says I'm wrong. I even worked backwards the answer in the book and it makes no sense to me. Please work this out and tell me what I'm doing wrong.

    1. The problem statement, all variables and given/known data

    A block of mass m1 = 250g is at rest on a plane that makes an angle of theta = 30 with the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of mass m2 = 200g that hangs freely by a string that passes over a frictionless, massless, pulley. When the second block has fallen 30.0 cm, what will its speed be?

    2. Relevant equations

    Fnet = ma (mass is in kg) ; v(final)^2 = v(initial)^2 + 2a(change in x)

    kinetic friction = uk Fn ;

    3. The attempt at a solution

    I started by drawing the first block. I changed the axes so that the surface of the incline would be the x axis and the y axis would be the direction of the normal force. So, the forces here should be kinetic friction, tension, normal force, and gravity. I then split gravity (which is .25 * 9.8, or 2.45 N) into its x and y components. Since the incline is 30 degrees, the force of gravity makes a 60 degree angle with its x component. Thus, the x value is 2.45cos(60). And the y value, 2.45sin(60). Thus, the normal force is 2.45sin(60), or 2.12N. I then calculated the magnitude of kinetic friction, which is .212N. I will also mention at this point that the x value of gravity was 1.23 N. Next, I got the force of gravity for the second block, 1.96 N. I think this is the magnitude of the tension of the first block. I then used newtons second law to get a = 2.07 m/s^2. But I think I must misunderstand tension, because I added the friction force and x value of the gravity force for the first block to get what the tension force on the other block should be. I then used newtons law again for the second block and got a different acceleration. Either way, I get a different velocity than the one it says in the book. And yes, I used all kilograms and meters.

    Summary: I'm going crazy, please help.

    EDIT: Why does that template copy when you preview? Whatever.
     
  2. jcsd
  3. Feb 14, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    The mg of the 2nd block is not the tension force. See this by applying
    sum of forces = ma to the 2nd block.
    mg - T = ma, so T = mg - ma. NOT just mg.

    The two blocks and the rope are one system with the same acceleration. Apply
    sum of forces = ma
    to the whole works. Take that mg as positive, the component of force down the ramp and the friction as negative because they oppose the motion. Use the combined mass in the ma part.

    Once you know the acceleration, just use accelerated motion formulas to find the speed.
     
  4. Feb 14, 2009 #3
    Thanks a lot.
     
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