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Vertical Work done

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A string pulls a 50kg block 10m up and the acceleration of the block is g/10. What is the kinetic energy of the block?


    2. Relevant equations
    W = Fd
    W = 1/2*mv^2

    3. The attempt at a solution
    1st attempt (each force work separately)
    The work done on the block from the string:
    W_s = F_s * 10m

    F_s = ma
    F_s = 50kg * g/10 m/s^2

    W_s = (50kg * g/10 m/s^2) * 10 m = 490 J

    The work done on the block by gravity:

    W_g = F_g * -10m

    F_g = mg
    F_g = 50kg * 9.8 m/s^2

    W_s = (50kg * 9.8 m/s^2) * -10m = -4,900 J

    If I add the net Work done up (-4410 J), the answer is incorrect.

    2nd attempt (net forces first)

    W_s = F_s * 10m

    F_s - mg = ma
    F_s = mg + ma
    F_s = (10*9.8) + (10*[g/10]) = 107.8 N

    W_s = 107.8 * 10m = 1078 J

    -mg + F_s = ma
    mg = F_s - ma
    mg = 107.8 N - (10 * [g/10])
    mg = 98
    g = 98/10 = 9.8

    W_s = 9.8 * 10 = 98 J

    1078 + 98 = 1176 J

    Both of the answers should be the same but they're not, and both answers are incorrect.

    The first method was not using the net force and using each force separately to find the work done then add them together, and the second was finding the net force in the up direction and then the down direction and then adding them together/
     
  2. jcsd
  3. Oct 25, 2012 #2

    CWatters

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    Is that the exact wording of the problem?

    If so I note that it asks only for the KE. So you have a block that's accelerated at g/10 for a distance of 10m. You only need to know how fast it's going at the end of the 10m using standard equations of motion.

    No need to calculate the work done on the block by the string or gravity or it's PE.
     
  4. Oct 25, 2012 #3

    tiny-tim

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    Hi PhizKid! :smile:

    I don't really follow either method. :confused:

    m is 50, and I don't see a 50 anywhere.

    And why are you separating the forces? The kinetic energy equals the total work done, which is the total force "dot" distance …

    so just use the total force.
     
  5. Oct 25, 2012 #4
    Oh, I made some substitution errors. But in any case, we are working on Work and KE in class so I want to stick with those for practice. The prof also said that we can take the work done in each direction and then add them up in the end which is what I attempted in the first try. The second time I took the work for the net force going up and the work for the net force going down since there is an acceleratio going up. So no constant velocity
     
  6. Oct 25, 2012 #5
    This:
    3. The attempt at a solution
    1st attempt (each force work separately)
    The work done on the block from the string:
    W_s = F_s * 10m

    F_s = ma
    F_s = 50kg * g/10 m/s^2

    W_s = (50kg * g/10 m/s^2) * 10 m = 490 J

    Gives you the work done by the net force, or the total work done since the block is accelerating under the influence of the resultant force.

    The work done by gravity is -490 J. You made an error in the calculation.
     
  7. Oct 25, 2012 #6
    I just got back from my physics class. The solution the professor gave us was:

    1) Work done by the force pulling the string:

    W_s = F*d

    F - mg = ma
    F = mg + ma
    F = (50)(9.8) + (50)(g/10)
    F = (50)(9.8) + (50)(9.8/10)

    W_s = F*d = [(50)(9.8) + (50)(9.8/10)] * 10
    = 5,390 J

    2) Work done by gravity:

    W_g = -F_g*d

    F_g = mg = (50)(9.8)

    W_g = -[(50)(9.8)] * 10
    = -4,900 J

    3) Total work done:

    5,390 J + -4,900 J = 490 J
     
  8. Oct 25, 2012 #7
    I think it made sense to me in class but now that I look over it again, I feel like something is wrong. Why is the total work done the work of the net force plus the work of the force done by gravity? Doesn't the work of the net force cover all of the work including gravity?
     
  9. Oct 25, 2012 #8
    F is not the net force it is the force pulling the string.
    The net force is F - mg.
     
  10. Oct 25, 2012 #9

    CWatters

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    In step 3) I think he means "Total Work done in both direction".

    I think he's trying to give you a very "full" explanation...

    In step 1) he's showing you how to calculate the total work done by the string ...

    W_s = Work done against gravity + work done accelerating the mass
    W_s = [(50)(9.8) + (50)(9.8/10)] * 10

    However to calculate the KE gained by the mass you need to subtract the work done against gravity (or add the work done by gravity) to leave only the work done accelerating the mass.

    It step 2 he basically repeats the calculation for the work done by gravity...

    W_g = Work done by gravity
    W_g = -[(50)(9.8)] * 10

    It's -ve because he said "Work done by gravity"

    Then finally he does the addition.

    W_s + -W_g = W_ke

    at least that's my interpretation of what he's done.
     
    Last edited: Oct 25, 2012
  11. Oct 26, 2012 #10
    Ok, so the work done by the string is the work done against gravity + work done by the accelerating mass

    Then why isn't the work done by gravity the work done against the string + work done by gravity?

    If the net force is F - mg, then F = mg + ma
    If the net force is F - mg, then -mg = ma - F

    So how come the work done by gravity isn't W_g = (ma - F) * 10m?
     
  12. Oct 26, 2012 #11
    The displacement is upwards, but gravity is acting downwards, so its work is negative, otherwise I am not sure why you want to make it so complicated. Work is a scalar so you just add up the quantities - the work done by all of the forces - which is what was done in 3).
     
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