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Homework Help: Find the velocity of the 14kg mass just before it hits the ground

  1. May 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A 14kg mass is attached to one side of a vertical pulley and an 8kg to the other.

    The 14kg mass is 5m above the ground. The 8kg is just resting on the ground. The pulley is frictionless and weightless.

    Find the velocity of the 14kg mass just before it hits the ground.

    2. Relevant equations

    v^2=u^2+2a*s where s is the distance

    3. The attempt at a solution

    Draw a free body diagram for the 14kg mass. Take downwards as positive,

    Let g= acceleration due to gravity. The the net force on the 14Kg body is
    F=14g-8g Newtons (acting down).

    The 14Kg accelerates downwards with acceleration, a, given by:

    a=(14g-8g)/14 = 4.2 m/s^2

    Thus, since the acceleration is constant:
    v^2=0 + 2*a*5

    v= 6.5 m/s

    This doesn't agree with the answer given is 5.2 m/s^2! So where have I gone wrong?
  2. jcsd
  3. May 9, 2014 #2
    Your calculation for the acceleration is incorrect. I'd take a step back and spell out exactly how you're deriving your force equation. Remember there is an acceleration constraint. There will be at least two separate equations concerning the net forces on the heavier mass and the lighter mass. You'll need to relate those equations to derive a single equation for the acceleration.

    Addendum: there will be three separate equations if you also want to spell out the tension in the rope holding the masses together. Notice the rope's tension will be constant given the wording of the problem.
  4. May 10, 2014 #3
    Oops. Seems I skipped a step! Thanks.

    => (8+14)a= 14g-8g
    which gives the correct answer after a bit of algebra.

    If the down direction is taken as positive, shouldn't these equations be...

    (T1 act upwards

    which gives a= (14g+8g)/(14g+8g) which is wrong.

    Where is the error in the signs?
  5. May 10, 2014 #4


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    You're forgetting that the pulley turns the string 180 degrees. The tensions are the same, including the sign, as far as the weights are concerned.
  6. May 10, 2014 #5


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    I don't quite understand why you have to worry about tension and all these redundant equations, your original line of reasoning was much more direct, and perfectly correct (except for one small error).

    You had everything right except you forgot that it's not only the 14kg side that's accelerating, it's the entire system (so the mass would be (14+8)kg). So instead of dividing the force by 14 to get the acceleration, you should have divided by 22.

    This would have given you the correct answer.
  7. May 10, 2014 #6


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    PeterPeter figured that out already (see post #3).
    But he'd like to understand where his second approach goes wrong.
  8. May 10, 2014 #7
    Obviously one doesn't have to for this particular problem, but doing so helps minimize errors for one; note that OP made his "small error" after skipping a step (as he admitted). Also, some physics professors require that you draw force diagrams, identify all forces (including tension) and write out corresponding equations, and otherwise show all work. Finally, sooner or later he'll run across tougher versions of this problem (in which there is more than one pulley, the pulleys have mass and/or friction, etc.), and keeping accurate track of all forces will become a virtual necessity.
  9. May 12, 2014 #8


    Here are my current thoughts.

    Taking down as positive:

    For 8Kg weight, where a is the acceleration of the 8Kg:

    For 14Kg weight where b is the acceleration of the 14Kg:

    For Pulley (since the tension on both sides of the string acts down):

    Finally, since one weight accelerates up and the other down:

    I guess (if the rope is massless etc)

    Is this correct so far?
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