1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

  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.


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

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

    a = 6.065 \frac{m}{s^2}

    v_f^2 = v_o^2 + 2a\Delta x

    v_f = \sqrt{2 * 6.065\frac{m}{s^2} * .47m}

    [tex] v_f = 2.388 \frac{m}{s} [/tex] And that was correct! Just forgot to multiply by the coefficient of friction, thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook