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A block of mass m1 = 2.40kg is connected to a second block of mass m2 = 1.80kg

  1. Feb 23, 2010 #1
    A block of mass m1 = 2.40kg is connected to a second block of mass m2 = 1.80kg....

    A block of mass m1 = 2.40kg is connected to a second block of mass m2 = 1.80kg, as shown in the figure. The two masses start from rest and are moving with a speed of 2.05 m/s just before m2 hits the floor.

    http://i85.photobucket.com/albums/k80/krystek01/5416308036.jpg

    a. If the coefficient of kinetic friction is mu k = 0.350, what is the distance of travel, d, for the masses?

    b. How much conservative work was done on this system?

    c. How much non conservative work was done on this system?


    Please help I just can't get this problem right. If someone could walk me through it and show me step by step how to do it, I will greatly appreciate it. Thanks :)
     
  2. jcsd
  3. Feb 24, 2010 #2
    ok so here it goes:

    U initial = m1gh + m2gd
    K initial = 0
    E initial = Ui +Ki = m1gh + m2gd

    U final = m1gh + 0
    K final = .5m1V^2 + .5m2V^2
    E final = m1gh + .5m1V^2 + .5m2V^2

    Wnc = -coeff. of friction x m1 x g x d
    Wnc = E final - E initial

    I tried to solve it for d, and what i got is not the solution.
    I just tried the energy conservation that Ef = Ei of the system, and solve it for d, again wrong. There is really no other force that you can apply since its about conservative and nonconservative energy. And to calculate those you need d.
     
  4. Feb 24, 2010 #3

    rl.bhat

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    Homework Helper

    First of all you have to find the acceleration of the blocks.
    Draw the free body diagram for m1 and m2. If t is the tension in the string, then
    T - mu*m1*g = m1*a....(1)
    m2*g -T = m2*a.......(2)
    Solve these two equations and find a.
    Final velocity of m2 is given. Accelerations is known. Using kinematic equation find d.
     
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