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Homework Help: Is more work done?

  1. Feb 22, 2008 #1
    This is not HW, just something I was thinking about...

    Let's say two peole each lift a one kilogram mass from the floor to a height of 1.0 m off the floor.

    If both people lift with a force of 9.8 N, the masses are lifted at the same constant speed, each does the same amount of work and each delivers the same amount of power (because equal times).

    If the exercise was conducted again and Person 1 lifted with a force of 9.8 N, Person 1 would do the same amount of work as in the first scenario and deliver the same power. But, if Person 2 was to raise the mass to a height of 1.0 m in a shorter time than Person 1, Person 2 would have delivered more power.

    Did Person 2 also do more work? Person 2 would have had to apply a greater force than the 9.8 N, resulting in more force over the same distance. More work, right?

    Is the work done against gravity equal to mgh, only if the object is lifted at a constant speed?

    Is it even possible for Person 2 to deliver more power in this scenario without doing more work?
  2. jcsd
  3. Feb 22, 2008 #2
    You've got most of this right, but I think you might be missing a couple of considerations. First, keep in mind that power is a rate, i.e. the rate at which energy is expended or work is done. So, for your last question, yes, someone doing the same work in half the time does so with twice the power.

    I think the thing that is tripping you up the most, however, is your intuition about the lifting force. A force of exactly g (~9.8 m/s^2) simply cancels gravity, so you get no motion. If you apply a constant force that is greater than g, then you'll get motion, but it's not with constant speed, since you now have a non-zero upward acceleration. In this case, the guy lifting with the greater force will indeed lift the mass faster -- because he's accelerating it at a higher rate -- but that also means that it will arrive at the 1.0 m height with a greater velocity, and thus with more kinetic energy. That's what's missing in your energy balance. The same amount of work is done against gravity in all cases (given by the change in gravitational P.E.), but more more work is done by accelerating the mass faster and thus giving it more K.E.
  4. Feb 22, 2008 #3
    Okay, I think I'm seeing it.

    So Person 2, by lifting it faster does do more work, but the extra work he does serves increases the KE of the mass, not overcome the work gravity does.

    Whereas, the work Person 1 does is used to overcome the work gravity does.

    For Person 1, the net work zero? Person 1 does pos. work, gravity does neg. work

    For Person 2, the network is not zero because of the greater KE? So the net work is the KE that the mass has at the 1.0 m mark.
  5. Feb 22, 2008 #4
    Well, you have to be a little more careful in stating the problem. I believe you suggested that these guys are applying constant forces, i.e. they're not "slowing down" as they reach the 1 m mark. In that case, both guys do the same work overcoming gravity, but different amounts of work accelerating the mass, as exhibited by the different final velocities.

    If you are thinking that they do "slow down" and bring the mass to rest at 1 m, then you have to consider the changing acceleration, which gets more complicated. Even guy #1 is still accelerating his mass to some non-zero velocity, however, so he's given it some K.E. for at least a short while and therefore he will have done a little more work than just what is given by the change in P.E.

    No matter what, though, both these guys will do positive work. The minimum work they do will be equal to the change in P.E. (you know this just from conservation of energy). I guess the limiting case would be where a third guy accelerates his mass to a tiny upward velocity, then maintains constant velocity by setting his upward force to equal the gravitational force exactly (so the mass is just moving inertially), and then lets gravity decelerate the mass at the 1 m mark. If you let this small velocity get arbitrarily small (so the time goes toward infinity), then the added K.E. get correspondingly small, and the work done approaches the change in P.E.

    No one ever talks about it this way, though! That's because for practical purposes, the change in P.E. is all that is relevant, since that's the only energy that can be recovered when you simply move a mass from one height to another - the K.E. it had while in motion is gone. You brought in these considerations by asking about the time taken to move from one height to the other, so I think that's where the K.E. has to be considered.
  6. Feb 22, 2008 #5
    Great responses. Thank you. You've definitely cleared it up for me.
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