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Physics. Linear Mechanics question

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 5.00kg hanging block is connected by a string over a light pulley to a 3.00kg block that is sliding on a flat table. The coefficient of friction between the 3.00kg block and the surface is 0.400. The system starts from rest. What is the speed of the 5.00kg block when it has fallen 1.50m?

    x1 = 0
    x2 = 1.50m

    Block 1

    m1 = 5.00kg

    Block 2

    m2 = 3.00kg
    coefficient of friction = 0.400

    2. Relevant equations

    Block 1
    x: Fnet = T1 + Tf = ma
    y: Fnet = Fg + N = ma = 0

    Block 2
    x: Fnet = ma = 0
    y: Fnet = T2 + Fg = ma = T2 - mg = ma

    3. The attempt at a solution

    a = (t1 + tf) / m

    a = (t2 + Fg) / m

    (t1 + tf) /m = (t2 + Fg) / m

    Plug block 2 (x-axis) t2 = ma + mg into block 2 equation

    ma + mg + T1 = ma

    mg + T1 = ma - ma = 0

    T1 = -mg

    I don't know if this is right. Does that mean the tension on the sliding surface is equivalent to the negative Fg?
     
  2. jcsd
  3. Sep 8, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    (1) It looks like you reversed the blocks. Block 1 was the hanging block.
    (2) There's only one tension in the rope; just call it T.
    (3) For the sliding block, which direction do the friction and tension forces act?
    (4) For the hanging block, which direction do the tension and gravity forces act?
    (5) Since the masses are different, label them differently: m1 & m2.
     
  4. Sep 8, 2009 #3
    1. Just noticed.

    2. Acknowledged.

    3. Friction on the sliding block acts in the opposite direction of tension.

    4. Tension acts in the opposite direction of gravity.

    5. Will do.

    Thanks. :)
     
  5. Sep 8, 2009 #4
    Is the distance, deltaX = 1.5, negative since the hanging block descends in height?

    Thanks.
     
  6. Sep 8, 2009 #5
    One more question. Instead of equating the acceleration of the two equations, I isolated the tension.

    m1 = 5.00kg
    m2 = 3.00kg

    u = coefficient of friction

    Block 1
    T + Fg = m1*a ---> T + (-mg) = m1*a ------> T - mg = m1*a ----> T = m1*a + mg

    Block 2:
    T + Ff = m2*a ---> T + (-uN) = m2*a ---> T = m2*a + uN

    m1*a + m1*g = m2*a + u*N

    5.0*a + 5.0*(9.8) = 3.00*a + 0.400*3.00*9.8
    5.00a + 49 = 3.00a + 11.76
    2.00a = - 37.24
    a = -18.62 m/s^2

    Then plug in a into the displacement/ velocity equation: v^2 = v0^2 + 2as

    Is that correct?

    Thanks.
     
  7. Sep 8, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Careful with signs. The acceleration of block 1 will be downward, thus -a.

    OK.

    Sanity check: Compare your calculated acceleration with g. Do you think it's possible for the block to accelerate greater than g?
     
  8. Sep 8, 2009 #7
    So it would be

    Fnet = T + Fg = - ma
    T + (-mg) = - ma
    T - mg = - ma
    T = - ma + mg

    Correct?

    Acceleration of Fnet cannot exceed gravity downwards, because it is the difference between Tension and Fg.

    T + Fg = - m1*a ---> T + (-mg) = - m1*a ------> T - mg = - m1*a ----> T = - m1*a + mg

    The result, 4.655 m/s^2.
     
  9. Sep 9, 2009 #8

    Doc Al

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    Staff: Mentor

    Excellent!

    Here's a tip that might make things easier (although there's nothing wrong with what you did, just personal preference): Rather than isolate T, just add the two equations to make T cancel:

    (1) m1g - T = m1a
    (2) T - μm2g = m2a

    (1) + (2) gives:

    m1g - μm2g = m1a + m2a = (m1 + m2)a
     
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