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Work and Energy of block

  1. Mar 8, 2004 #1
    Question: The 20 lb block B rests on the surface of a table for which the coefficient of kinetic friction is 0.1. Determine the speed of the 10 lb block A after it has moved downward 2 ft from the rest. Neglect the mass of the pulleys and cords.

    Picture Attached

    Equation: T1 + Summation of work = T2

    This is what I got.

    s = displacement

    0 + [(Force of Friction)*(s) - (Weight of A)*(s) - (Weight of C)*(s) = 0.5*mA*v^2 + 0.5*mB*v^2 + 0.5*mC*v^2

    2*2 + 10*2 + 6*2 = 0.15528*v^2 + 0.31056*v^2 + 0.09317*v^2

    38 = 0.55901*v^2
    v = 8.24 ft/s

    Is it right?

    Attached Files:

  2. jcsd
  3. Mar 8, 2004 #2

    Doc Al

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    Staff: Mentor

    You made an error with the signs of the work done to lift/lower the blocks. Since one raises and one lowers, the signs must differ.

    To keep better track of signs, try thinking this way:
    ΔKE + ΔPE = -(Work done against friction)
  4. Mar 8, 2004 #3
    Thanks! What a silly mistake. The answer is 6.27 ft/s!
  5. Mar 9, 2004 #4

    Doc Al

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    Staff: Mentor

    I get a different answer. Recheck your numbers.
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