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Homework Help: Sliding block

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A 5.86 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 10.00 N at an angle = 35.0° above the horizontal (x-axis). What is the speed of the block 3.70 seconds after it starts moving?

    2. Relevant equations

    Fcos() - ma = 0

    Fsin() - mg = 0

    V_f = V_0 + at (V_0 = 0)

    3. The attempt at a solution

    Since the first two equations are equal to zero I set them equal to each other and then solved for acceleration. I then used this in the last equation to solve for V_f. Apparently I am doing something wrong because I am not getting the correct answer.
     
  2. jcsd
  3. Sep 10, 2010 #2

    rl.bhat

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    Homework Helper

    What is the value of the acceleration? Will you show your calculations?
     
  4. Sep 10, 2010 #3
    Re: Sliding block calculations

    Fcos() - ma = 0

    Fsin() - mg = 0

    so setting these two equations together and inputting given data:

    10cos(35) - 5.86(a) = 10sin(35) -5.86(9.81)

    8.192 - 5.86(a) = 5.736 - 57.487

    -5.86(a) = 5.736 - 57.487 - 8.192

    -5.86(a) = -59.943

    a = 10.22918089 m/s^2

    Now to solve for V_f (final velocity)

    V_f = V_0 + at

    V_f = 0 + 10.23(3.7) = 37.84 m/s

    FAIL!!!!!! This is the wrong answer
     
  5. Sep 10, 2010 #4

    rl.bhat

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    Homework Helper

    10cos(35) - 5.86(a) = 10sin(35) -5.86(9.81)

    This step is wrong. There is no acceleration in the vertical direction because the action and reaction are equal and opposite. Hence

    F*cos(θ) = ma
     
  6. Sep 10, 2010 #5
    Ok....got it

    This always holds true correct, except when the object is inclined/declined

    here are the proper calculations:

    10cos(35) = 5.86(a)

    a = 1.39 m/s^2

    V_f = 0 + 1.39(3.7) = 5.17 m/s

    correct answer

    Thanks a bunch
     
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