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What is negative mechanical Power?

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    In terms of mechanics, does negative Work produce negative Power? If so, what is negative mechanical Power?


    2. Relevant equations
    W=F*d
    P=W/t


    3. The attempt at a solution
    I know if a car is rolling downhill and I try to push against it, but I end up getting pushed backward, I'm doing negative Work because the car is moving in the opposite direction of the force applied against it. Does this mean I'm producing negative Power? If so, what does it mean in practical terms? (It's been 30 years since high school physics, and I'm really at the limit of what I remember.) Thanks!
     
  2. jcsd
  3. Oct 8, 2012 #2

    Simon Bridge

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    "Homework" means that how we can answer the question is a bit different...

    Think of work and power in terms of conservation of energy instead.

    If you are doing negative work - what are you doing that work "on"?
    What is happening to your energy?

    Note: in your example, the net force is pointing down the slope, and so is the displacement: so positive work is being done but what is doing the work?

    A better example: Superman rushes to stop an out-of-control train. He braces his feet against the tracks and his hands against the front of the locomotive and pushes hard ... he slides backwards on the tracks but the train slows to a stop just in the nick of time (Lois was tied to the track or something).
    What happened to the energy here?
     
  4. Oct 8, 2012 #3
    This is a bit out of my league, so I'm sure you'll point out if I'm on the right track here...

    I'm doing negative work against the car, but the car is doing positive work against me, since I'm being displaced in the same direction as the force being applied by the car?

    I'm sorry, I'm not sure I understand the question here. Some of Superman's energy is going into the train and some of the train's energy is going into Superman, with the net effect that displacement equals zero? I'm not sure why this is a better example of "negative power." In my mind, Superman is doing negative work because he's moving in the opposite direction of the force he's applying against the train. I can see he's doing negative work. I can see the change in how much negative work he's doing, depending on how much he's being displaced. Is his rate of negative displacement indicative of negative power?

    So let's say one time against the train he gets displaced 50 feet, and the second time against the same train traveling at the same velocity he gets displaced 10 feet. He performed more negative work when he got displaced 50 feet, right? Would this mean he also produced more negative power? If so, are we simply talking about the rate at which he gets displaced?

    So if he gets displaced 50 feet in 10 seconds one day, but gets displaced 50 feet in 5 seconds the next day (assuming he's applying the same force each day, of course), does this mean he produces more negative power when he gets displaced in 5 seconds? If so, does negative power simply refer to the rate at which negative work is performed?

    I feel like I might be on the right track, but I wasn't that great at high school physics all those decades ago...Thanks for your patience!
     
  5. Oct 9, 2012 #4

    Simon Bridge

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    The force doing the work on you via the car is gravity.
    In terms of energy - the car is losing gravitational potential energy and so are you... but you are both gaining kinetic energy.

    Your applied force is preventing kinetic energy being gained as fast as it would be otherwise.

    Well superman and the train do have a net displacement ... since superman's feet slide along the tracks.

    Work is change in some energy of the system ... implying you are not thinking of a closed system. Negative work being done on something means that energy is being taken away from it.

    eg. you push on something heavy, it accelerates, you lose chemical energy in your muscles and the object gains kinetic energy in it's motion. The magnitude of either energy change is the work done ... the negative for you means you did the work, and the positive means the object had the work done on it.

    It gets curly when you consider multiple things decreasing - you have to track double-negatives.

    Power is the rate that the energy is changing.

    It is not how we normally think about things though - you cannot build a machine to generate negative power for example.

    Also see:
    http://answers.yahoo.com/question/index?qid=20080704142322AAb0bGv
    http://en.wikipedia.org/wiki/Work_(physics [Broken])
     
    Last edited by a moderator: May 6, 2017
  6. Oct 26, 2012 #5
    Thanks so much for your reply. Sorry to take so long to respond (and express gratitude).

    In trying to look at these ideas pretty simplistically, your explanation of the ways in work (energy?) is being generated (or lost) is helpful. It sounds like I'm overly complicating things because of semantics (which was my downfall in high school physics as well).

    (1) Work is the change in energy, and power is the rate of energy change.

    (2) Positive work is when energy is added to an object, and negative work is when energy is lost from an object

    (3) Positive power is the rate at which energy is added to an object

    (4) Negative work is the rate at which energy is lost from an object

    So when I'm pushing against the car rolling downhill, the car is gaining energy because of kinetics and my push isn't enough to negate that gain in energy. So the car's power is measured by the rate at which kinetic energy is being added to the car.

    Likewise, when I'm pushing against the car rolling downhill, I'm losing energy because of a loss of kinetic energy (as well as loss of chemical energy?). So I'm experiencing negative power, which is just the rate at which this kinetic and chemical energy is being lost from me.

    Sorry for simplifying so much. It sounds like I was complicating things, and I just want to make sure I actually understand.

    Thanks again!

     
    Last edited by a moderator: May 6, 2017
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