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Mixed up with Work and Power and Force

  1. Oct 31, 2009 #1
    Hi,

    I have a mix-up with power and work and force too. My problems stem from this question:

    A plane of mass M consumes P Power in order to stay afloat. Express the power used in terms of M and other relevant constants.

    The more I thought about this problem, the more I got confused. It makes sense that you are doing work and consuming power because you are working against gravity. However, I remembered my teacher telling me that work is force times distance, so if you push a stationary object, no matter how much force you push it with, if it does not move, you have done no work.

    That is where my confusion comes in. The plane is not moving at all, so technically no work is done. If no work is done, there should be no power consumption (work done per unit time). If there should be no power consumption, why are you using power? And it makes sense that you are using power because you are doing work (or something else) against gravity.

    I tried it out and thought maybe you have to assume that the only force exerted on the plane is gravity. From the gravity, you can calculate the rate of change of distance by using the kinematics equations, and from that, you can work out the work done gravity. And the power consumption will be the negative of this value. However, this does not make sense, because the work done by gravity will increase as time passes (As time passes, the velocity will become greater, so the distance travelled will become greater and thus more work is done), but I think the power P is a constant value.

    So what should be done here? Do I have a fundamental flaw in my understanding of power, work and force? Is there a formula that describes the relationship between power and force without a length term? Thanks.
     
  2. jcsd
  3. Oct 31, 2009 #2
    For an object that floats, gravity is a downward force on it, buoyancy is an upward force on it, equal in magnitude, for a net force of zero. No displacement, no work, no power.
     
  4. Oct 31, 2009 #3
    Post # 2 is the way I would answer your question as well... For a confirmation from a different perspective, work is change in energy, read the introduction here:

    http://en.wikipedia.org/wiki/Work_(physics)
     
    Last edited by a moderator: Nov 1, 2009
  5. Oct 31, 2009 #4
    i'm having the same trouble as you, it's a hard concept to grasp fully =[

    but physically you can't help but think that in order to stay motionless it must be accelerating gas or something beaneath it (like a rocket), giving them an energy of 0.5mv^2 etc but i suppose that still wouldn't be classed as doing work...
     
  6. Oct 31, 2009 #5

    russ_watters

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    Staff: Mentor

    Note that the difference between input power and output power is efficiency. For an object such as a helicopter (plane?) that is hovering due to an input of power, the output power and thus efficiency of the system as measured by motion of the helicopter is zero.

    But really, how is that different from most powered vehicles? Unless a car or a horse (heck, even a pump, often times) is lifting an object to a higher gravitational potential, all of the power used is typically wasted as frictional loss. If you drive your car to work and then back home, your car has ended up exactly where it started and burned a bunch of fuel to ultimately achieve nothing but fighting friction.
     
  7. Oct 31, 2009 #6
    When posting a link that ends with the closing of parentheses, this forum always inserts the "slash url" bbcode before the right-hand parenthesis instead of after it, making the link fail to work, unless you edit it manually.
     
    Last edited by a moderator: May 4, 2017
  8. Oct 31, 2009 #7
    Power = force X velocity
     
  9. Oct 31, 2009 #8
    People usually ask the question this way: let your arms hold up some heavy weights for a long time - if you aren't "doing any work", why do you get tired out? It's because you aren't doing any work on the positioning of the weights, but you are doing work on something else. You are doing work on the molecules in your own body, making them undergo chemical changes. To refer to work implies that you have to consider exactly what object is the recipient of that work, and it's not the motionless object that you're paying attention to.
     
  10. Oct 31, 2009 #9
    To mikelepore: I am assuming that the air does not exert any buoyant force on the object. Something in the object itself is exerting a force the same magnitude as gravitational force upwards, so the object is stationary.

    To Naty1: Your link does not seem to work :(. Anyway, I know the work-energy therem and that work done is equal to the change in kinetic energy, but kinetic energy did not change in this case, and yet some work is done. (Ok, I fixed the link)

    To Chewy0087: Are you talking about assuming rocket propulsion? However, what happens if I am talking about a general case? It can be rocket propulsion (with minimal change in mass of the object), it can be a helicopter, it can be a plane flying horizontally at the optimal speed such that the difference in pressure allows it to maintain its height.

    To russ watters: For your example, the direction of the force changes, and since friction is non-conservative, you cannot just take the start and end points. You have to take the path. But I am talking about a conservative force now, and the direction of the force did not change, so... :(

    To mikelepore: Thanks. I read the part about 0 work. It says that if you put a book on the table you are not doing any work because there is no transfer of energy. But then, as you mentioned, when you push on the object, you do consume energy.

    You say that work is velocity * force. But I cannot use this equation in this case because it is not moving :(.

    As for your explanation about doing work on the molecules, I think you are saying that we are doing work by changing chemical energy to mechanical energy. But if we are not doing work, where does this energy go to? Also, if we think about it this way, how do we calculate the power? Are you saying that it cannot be calculated?

    Ok, I am getting more and more confused :(.
     
  11. Nov 1, 2009 #10
    Yes of course, work which is the change in KE over two distances(as I found out =P) takes account of conservative forces such as GPE as a force. yes, you're using your engine (and electric power), be it rocket propulsion or accelerating a volume of air beneath you etc in order to stay stationary, but no overall work is being done because 1) there is no net force on it & 2)it's not being moved.
     
    Last edited: Nov 1, 2009
  12. Nov 1, 2009 #11

    rcgldr

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    I assume you mean an airplane in level flight? The power consumed is equal to the drag force times the air speed. The lift to drag ratio at some stated speed and the weight of the plane determines the drag force: drag_force = weight * (drag / lift), in level flight.

    As an example of an efficient aircraft, a Nimbus 4T glider has about a 60:1 glide ratio at about 68 mph. Assuming that it can acheive this same ratio with forwards speed of 60 mph, it's downward speed would be 1 mph, and a typical weight for the glider would be 1500 lbs. Gravity is supplying the force, and the power consumed equals the weight times the descent velocity. In this case, the power = 1500 lbs * 1 mph / 375 (conversion factor) = 4 horsepower.
     
  13. Nov 1, 2009 #12

    russ_watters

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    Take the whole path if you want - net useful work is still zero! All you've done is fight friction and waste all your energy.
     
  14. Nov 9, 2009 #13
    Can a plane be float without moving at a certain height?Then equating the bouant force and weight of the plane downward, the density of plane=that of air!!!!!!!Is it physically possible?Physically interpreted data should be that the density of plane>that of the air medium.So to maintain a certain position of a plane,a certain velocity is required.....at least I think so.
     
  15. Dec 9, 2009 #14
    The key fact here is that power is being used to maintain the plane in a stationary position, suggesting that the plane will move or fall without the power. What happens quantitatively if you suddenly stop supplying the power. The plane falls at a certain rate depending on the force of gravity, as well as the air resistance, the bouyant force, the wind, and any other force acting on the plane. This is the source of the mathematical expression that you are looking for.

    The power that is being supplied to maintain the plane in its stationary position is equal in magnitude to the rate at which energy would be manifested due to the net force on the plane without the power input. Relatively speaking, the work that is being done is the net force times the distance that the plane moves if the power is no longer supplied.

    Furthermore, suppose the plane is falling out of the sky. In its frame of reference it is not moving. Does that mean no work is being done? Of course not.
     
  16. Dec 9, 2009 #15
    When you say that no work is being done because the plane is not moving, what you are really saying is that no NET work is being done which is not the same as no work at all being done. It's like saying that if the net force on it is zero, then there are no forces at all. This is not the case. Forces are acting on the plane. Work is being done. The net is zero. This gives you an equality to work with.
     
  17. Dec 9, 2009 #16
    Please refer to the four forces acting on an airplane at this link:

    http://www.grc.nasa.gov/WWW/K-12/airplane/forces.html

    Weight = mg

    For lift to equal weight requires forward velocity v. Lift equation:

    http://www.grc.nasa.gov/WWW/K-12/airplane/lifteq.html

    Velocity v creates drag:

    http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

    To overcome drag requires thrust T. The power of thrust is T times v:

    P = Tv

    You should be able to present a respectable effort at a solution from this information.
     
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