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Use Newton's second law to find the net force acting on both blocks

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Two blocks with masses m1 = 1.10 kg and m2 = 3.20 kg are connected by a massless string. They are released from rest. The coefficient of kinetic friction between the upper block and the surface is 0.490. Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 47.0 cm.

    Hint: Use Newton's second law to find the net force acting on both blocks. Since they are connected by a string, they act as one body and have the same acceleration.

    3. The attempt at a solution

    So I did that, but ended up wrong.

    [tex] \Sigma F = ma [/tex]

    [tex] -f_k + mg = ma [/tex] Because the tension forces would cancel.

    [tex] \frac{-(1.1kg * 9.8\frac{m}{s^2}) + (3.20kg * 9.8\frac{m}{s^2})}{(1.1kg + 3.2kg)} [/tex]

    [tex] a = 4.786 \frac{m}{s^2} [/tex]

    [tex] v_f^2 = v_o^2 + 2a\Delta x [/tex]

    [tex] v_f = \sqrt{2 * 4.786\frac{m}{s^2} * .47m} [/tex]

    [tex] v_f = 2.12 \frac{m}{s} [/tex]

    But this was wrong, so where did I go wrong?
     
  2. jcsd
  3. Sep 23, 2009 #2
    Re: Pulley

    It looks like you had friction in your equation (-fk) but did nothing with it. I think you calculated the velocity of the blocks without any friction.

    To be honest, I don't really understand the problem. What does it mean by 'between the upper block and the surface'?
     
  4. Sep 23, 2009 #3
    Re: Pulley

    Well we had a picture, and I don't know how to transfer it over. Basically m1 is on a surface with a string attached to m2. m2 isn't on a surface. The only forces acting on m2 are mg pulling it down and the tension pulling it up.

    Think of a cliff, where m1 is on the cliff and m2 is not on the cliff.

    Edit: Aha! found a picture

    http://spiff.rit.edu/classes/phys311/workshops/w5b/level_atwood/cart_and_weight.gif
     
  5. Sep 23, 2009 #4
    Re: Pulley

    Yeah, that was my assumption. = ), just wanted to be clear, especially that the upper block=m1.

    Try finding the force of friction applied to m1.

    Fk=uk*FN

    This is the force applied opposite to the direction of m1's motion.
     
  6. Sep 23, 2009 #5
    Re: Pulley

    [tex]
    \frac{-(1.1kg * 9.8\frac{m}{s^2} * .49) + (3.20kg * 9.8\frac{m}{s^2})}{(1.1kg + 3.2kg)}
    [/tex]

    [tex]
    a = 6.065 \frac{m}{s^2}
    [/tex]

    [tex]
    v_f^2 = v_o^2 + 2a\Delta x
    [/tex]


    [tex]
    v_f = \sqrt{2 * 6.065\frac{m}{s^2} * .47m}
    [/tex]

    [tex] v_f = 2.388 \frac{m}{s} [/tex] And that was correct! Just forgot to multiply by the coefficient of friction, thanks!
     
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