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Homework Help: Simple Problems involving Work

  1. Mar 9, 2010 #1
    1.1. You drop a ball from a height of 2m above the ground. Describe what happens to its potential energy, kinetic energy, and total energy as the ball falls towards the ground.

    2. Estimate (to the nearest order of magnitude) how much work I must do to pick up a 12 kg box to a height of 1.3m. Does it mattert how I pick up the box (fast or slow, straight up or along a curvy path, etc.)?

    3. In which of the following cases do I do positive work on a ball?

    I pick up a ball from the ground to a height of 1.4m
    I slowly set a ball down from a height of 1.4m down onto the ground.
    I hold a ball still at height 1.4m.
    Two of the above
    All of the above

    4. Explain your answer to the multiple choice question above.

    2.U=mgy W(grav)=U(i)-U(f)

    3. 1. The potential energy is the highest at the top (2m above the ground) because that is when it is highest. As it falls and picks up speed, though, the potential energy gets transfered into kinetic energy, and kinetic energy increases, and potential energy decreases. The total energy, though, is the same throughout the process.

    2. 12*9.8*1.3=152.88 J (nearest order of magnitude 150 J)

    It doesn't matter how you pick up the box, as long as the displacement is the same. For the gravitational work, it doesn't matter how fast or slow you pick up the box, but for total work, where kinetic energy is included, then, the difference between starting and ending velocities would be important.

    3. I slowly set a ball down from a height of 1.4m down onto the ground.

    4. "I slowly set the ball down from a height of 1.4m down onto the ground" is the only option that includes reducing the height, which would cause the gravitational potential energy to decrease, and for gravitational force to do positive work on the ball.

    Thanks for the help guys!!!
  2. jcsd
  3. Mar 10, 2010 #2
  4. Mar 10, 2010 #3
    I think you have explained very well. Except that you weren't answering the questions in 3 and 4, it asks which one of those involves YOU doing POSITIVE work.
  5. Mar 10, 2010 #4
    I think what I had is what the professor meant. Otherwise, I don't think any of the options would work.

    EDIT: Or it could be the first option, but that would mean increasing Gravitational Potential Energy, which is negative work.
    Last edited: Mar 10, 2010
  6. Mar 10, 2010 #5


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    No, the professor meant to ask which of the scenarios involve YOU doing positive work. You said "gravitational force to do positive work on the ball.", but the question asks about you doing positive work, not gravity.

    W=F dot d, so for W to be positive, F and d have to be in the same direction. That's not the case for "I slowly set the ball down" because the hand is pushing up while the ball is moving down.
  7. Mar 10, 2010 #6
    Well, I guess it is the first option then. The force from your hand is pushing the box up, and the displacement is up, as well. Thus, that option involves it doing positive work. Now, setting it down would count, as well, right? You're using a force from your hand to put the object down, and displacement is down, as well.
  8. Mar 10, 2010 #7


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    No, you're using the force from your hand to push the object up, so that it moves down slower than it would in freefall.
  9. Mar 10, 2010 #8
    Thus, reducing kinetic energy?
  10. Mar 11, 2010 #9
    OK, let me try this again. The box being lifted, because it has a lifting force that is in the same direction as the displacement, it has positive work being done on it by YOU. Now, when I set the box down, don't I still do positive work on it, because the force I am exerting on it from top to bottom is in the same direction as the displacement.
  11. Mar 11, 2010 #10
    Or would putting something down be considered negative work because you're increasing your potential energy to lift something?
  12. Mar 11, 2010 #11
    I am not sure what you meant by this. But this question is talking about "work done on the ball", not you.
    When you set something down, what is the direction of the force you are exerting on the object? Think about it. It is always up. It doesn't matter if you are supporting the object from the bottom up or if you are holding the object from above.
  13. Mar 11, 2010 #12
    OK, that clears things up. The force is constantly pushing up, while the displacement is down. The other option has both force and displacement in the same direction, though. These questions are on the early part of the chapter, and I read ahead. A sort of forgot the basics and I was thinking too deeply about gravitational and elastic energy, along with potential energy and what not. Not necessary for this. Thanks again, though!
  14. Mar 12, 2010 #13
    Sorry to bug you guys again, but I just want to make sure, was my answer for problem 2 correct?
  15. Mar 12, 2010 #14
    I think your answer is correct, although it seems ambiguous.
    1. If you raise an object to a height of H with an ending velocity, it will reach a height bigger than H. Therefore, I think to raise an object to a height H, it means that the object has an end velocity of 0.
    2. Starting velocity? I thought it should be assumed that the object was at rest at t=0. If you meant the speed after you have accelerated the object after a very short time interval during the lift, it shouldn't matter. You will do negative work on the object at the end to slow it down, if the object were not to have an ending velocity.

    So, if my arguments are valid, the work done on the object would be the same.
  16. Mar 13, 2010 #15
    Yeah, you're right. In this case, both the starting and ending velocities would be zero, so the kinetic energy differences between initial and ending kinetic energies would be zero (since they're both zero).
  17. Mar 15, 2010 #16
    If the starting and ending velocities were zero, wouldn't the total work be zero? The change in kinetic energy would be zero, and the work done by external forces on the object is equal to the change in kinetic energy.

    EDIT: That wouldn't mean that my answer for #2 should be zero, right? Because that's only asking about work that I do, and not total work. In the case of this equation K(f) + U(grav, f)= K(i) +U(grav, i) +W(other) The work that I do, in this case, would count as W(other), right? And it would be equal to U(grav,f) considering that K(i) and K(f) would both be zero, and U(grav, i) would be zero because that's at the bottom height (zero). BTW, the professor did confirm with me that initial and ending velocities are zero. Any help guys would be greatly appreciated.
    Last edited: Mar 15, 2010
  18. Mar 15, 2010 #17
    Right, good work!
  19. Mar 15, 2010 #18
    Thanks, I really appreciate it!
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