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Work done

  1. Dec 13, 2013 #1
    I've always thought of work done as being calculated by something like ∫F.dx whenever a force acts on a moving body. Another way of writing this ∫F.vdt. The non integral form of this equation (W = Fv) makes sense to me in the situation where the body is moving at a constant velocity as the driving force is doing work against some constant frictional force or whatever.

    But in the case where F is the only force acting (or where F is the resultant force) and the velocity of the body is changing then it would seem that the force does different amounts of work in different directions.

    Let's say that there's a body moving though space at a few km/s in some direction. If I apply a force in the direction of the velocity then it will do some work accelerating the body = ∫F.vdt. While this force acts the object moves though a large distance and so the work done is large.

    Whereas, if the same magnitude force acts perpendicular to the velocity for the same amount of time then the displacement in the direction of the force is small and therefore little work is done. Let's say that these forces are produced by burning fuel. To produce the same force for the same duration must require the same quantity of fuel. So if one force does more work, where does the energy go?

    I know that forces can act without doing work (e.g. supporting a weight against gravity) and that kinetic energy scales with v2, making small changes in the speed have a large effect on the kinetic energy when things are moving quickly. But it doesn't seem to make sense that burning the same amount of fuel to produce the same magnitude translational accelerating forces for the same durations should result in different amounts of work being done.

    Clearly something is wrong with my understanding! Could somebody make some suggestions on how I could clear this up?

  2. jcsd
  3. Dec 13, 2013 #2


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    I believe you need to look at the problem from an inertial frame at rest relative to the object before you apply the force. But I'm not 100% sure.
  4. Dec 13, 2013 #3
    You must include the energy transferred to the rocket exhaust to figure out where the energy goes.
  5. Dec 13, 2013 #4
    Could you elaborate on this at all?
  6. Dec 13, 2013 #5


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    When the rocket is slow, the exhaust gains more kinetic energy during burning.
  7. Dec 13, 2013 #6


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    The rocket exhaust forms a "plume" of spent fuel as it's ejected. Absent any external forces, the center of mass of the rocket and the spent fuel does not accelerate. In an inertial reference frame that is moving at the same speed as the rocket before any fuel is spent, then the center of mass of the rocket and spent fuel does not move.

    The total change of mechanical energy of the rocket and it's spent fuel will be the same regardless of what inertial frame is used, and is equal the amount of chemical energy in the fuel that is converted into mechanical energy by the rockets engine (most of the losses are related to heat).
  8. Dec 13, 2013 #7
    Yep, sounds good. But I'm still not seeing this clearly. Here's what I can't intuit...

    If I'm strapped into my seat in my spaceship's rest frame and then I switch on my two engines I'm going to feel a resultant acceleration that is the sum of the effects of the two engines. I get that after the burn my velocity in the original inertial frame will be √(Vx2 + Vy2). This gives me a change in kinetic energy of [itex]\frac{1}{2}[/itex] m (Vx2 + Vy2). If I look at my fuel gauge, I'd expect to have used fuel equivalent to this change in energy. I'd expect that both engines would have consumed the same amount of fuel and to each have added one half of this extra kinetic energy.

    Now lets say I start off working in some other inertial frame which gives some initial velocity A in the x direction. Now I get that after the acceleration I have a velocity of √[(A + Vx)2 + Vy2)] and a change in kinetic energy of [itex]\frac{1}{2}[/itex] m [(A + Vx)2 + Vy2)] which is A2 + 2AVx larger than the increase in energy as measured in the first inertial frame. So if we were calculating the fuel burnt in the second inertial frame then we'd get a different answer.

    So what am I doing wrong? Is it the calculations or some deeper assumption that I'm making?
    Last edited: Dec 13, 2013
  9. Dec 13, 2013 #8


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    From the frame initially at rest with the object:
    Let's assume a mass of 100 kg and an acceleration of 1 m/s2 for 10 seconds.

    Applying 1 m/s2 of acceleration for 10 seconds increases the velocity from zero to 10 m/s.

    Ke = 1/2 mv2, Ke = 1/2*100 (10*10), Ke = 1/2*100 (100), Ke = 1/2*10,000, Ke = 5,000 joules.

    Let's assume the fuel was a large mass (also 100 kg) ejected out the back and was accelerated an equal amount in the same amount of time. That leads to a velocity of 10 m/s in the opposite direction and a Ke of 5,000 joules also.

    Work done on ship: +5,000 joules.
    Work done on fuel: +5,000 joules.
    Total work done to fuel and ship: +10,000 joules.

    From the frame where the object is already moving:

    Let's assume a mass of 100 kg and an acceleration of 1 m/s2 for 10 seconds. The mass initially moves at 100 m/s.

    Ke = 1/2 mv2, Ke = 1/2 *100(100*100) = 1/2*100(10,000) = 1/2 (1,000,000) = 500,000 joules
    Initial Ke = 500,000 j

    Applying 1 m/s2 of acceleration over 10 seconds increases the velocity to 110 m/s, so new kinetic energy:

    Ke = 1/2 * 100 (110*110) = 1/2 * 100 (12,100) = 1/2 (1,210,000) = 605,000 joules.
    Final Ke = 605,000 joules.

    So the difference between the 2 kinetic energies is 105,000 joules.
    Work done on ship: +105,000 joules.


    Initially the fuel is identical to the ship and is moving at 100 m/s with 500,000 joules of Ke. After ejection, it is moving at 90 m/s in the same direction as the ship.

    Ke = 1/2 * 100 (90*90) = 1/2 * 100 (8100) = 1/2 * 810,000 = 405,000 joules.

    Difference in Ke = 95,000 joules.

    Work done on fuel: -95,000 joules (Work is negative because the force was opposite of the direction of motion)

    Total work done to ship and fuel: +10,000 joules (105,000 - 95,000)

    So what has happened here?

    This is called the Oberth Effect. Notice that in both cases the total work done is equal. HOWEVER, the work done to the ship is NOT equal. What has happened is that the fuel gives up some of its kinetic energy to the ship. In this case, the fuel gave up 95,000 joules of kinetic energy to the ship, with the remaining 10,000 joules coming from the chemical potential energy of the fuel before combustion.

    From wiki: In astronautics, the Oberth effect is where the use of a rocket engine when travelling at high speed generates more useful energy than one at low speed. The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy. The vehicle is able to employ this kinetic energy to generate more mechanical power.

    (Whew, this took me an hour to work out and to find the Oberth effect.)
  10. Dec 14, 2013 #9
    Thanks for the effort. I'd tried some similar calculations myself with actual numbers to get a better idea of what's up here. Including the reaction mass seems to make the numbers come out right. Now I'm trying to think of a situation where there is no reaction mass to worry about.

    What if we had a uniform g field and compared the work done by gravity for a free falling mass released from rest and a mass projected at some large initial horizontal speed? Let m = 1 kg and t = 1 s and g = 10. It seems that in any given time each mass would fall though the same gravitational potential gradient and therefore gravity does the same work.

    Working in the initial rest frame of the free falling mass gets us a change in kinetic energy of 50 J while the change in kinetic energy for the projectile is in the same frame is, err, the same. This is trivial but I think leads to a light bulb moment for me! Because if we now give the mass some initial speed parallel to the field we discover that the mass moves though a larger distance in the same time and since this takes the mass further down the potential gradient we can see that gravity has done more work!

    This is the problem that was bothering me at the start, how can a constant forces that act for the same time do different work. In hindsight, this is all implicit in the definitions that I started with. I trust someone will correct me if I'm wrong!
    Last edited: Dec 14, 2013
  11. Dec 14, 2013 #10


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    yeah, this is a much better thought experiment for the thing you were trying to logically reason about.

    yep. gravity does more work in the same amount of time when the object is given an initial downward velocity. And in both cases, gravity exerts the same force on the object.

    p.s. was there a similar thread like this recently?
  12. Dec 14, 2013 #11


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    I haven't done the math, but off the top of my head I'd expect that since more work is done on the mass, less work is done on the Earth since the mass has less time to act on the Earth. I would assume that the total work done is the same regardless of the velocity of the falling mass.
  13. Dec 14, 2013 #12


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    the work done over the same time (and with the same force) depends on the velocity of the falling mass. But the work done over the same distance (and same force) does not depend on the velocity of the falling mass.
  14. Dec 14, 2013 #13


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    Sure. Let's say a meteor is approaching the Earth. Wouldn't the gravity of the meteor displace the Earth a smaller distance the faster the meteor approaches, since there is less time for its gravity to act on the Earth?
  15. Dec 14, 2013 #14


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    MalachiK's thought experiment was with a uniform g field, no earth, which is good, since this is a nice and simple case. (or, it can be thought of as the limit when the earth is much more massive than the object, and the object only moves a small distance, compared to the radius of the earth).

    For the 'meteor directly approaching earth, with earth having zero initial velocity' problem, there is the added complication that the earth will move and the gravitational field depends on the separation of the two objects. I'm not 100% sure what your question is. Is it comparing two situations, where in both situations, the initial distance between earth and meteor is the same. But in one situation, the meteor has greater initial velocity (toward the earth). And in both situations, we let the asteroid move the same distance. And the question is then "will the earth have moved a different distance in these two situations?"

    it's an interesting question. need to have a think about that one. my instinct says yes it will be different.
  16. Dec 14, 2013 #15
    In the case of two similarly sized masses approaching from a large distance with some non-zero initial speed, isn't the work done by gravity just the same as the gravitational binding energy when the two masses come into contact with each other?
  17. Dec 14, 2013 #16


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    uh.. when I've seen the gravitational binding energy, it has meant the work required to separate the masses infinitely far away from each other, with zero initial KE and zero final KE, so I don't know how the gravitational binding energy would be defined when the objects have KE.
  18. Dec 14, 2013 #17


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    Alright, I don't know if this is what Drakkith was talking about or not. But I think I have the maths that says yes, the earth moves less when the meteor has greater initial KE. So, getting on with it: all you need is conservation of momentum. Due to conservation of momentum, and the fact that initial momentum is simply the mass of the meteor times its initial velocity, we get:
    [tex]X_E = \frac{M_m}{M_E} (X_m - vt) [/tex]
    and, to explain notation: ##X_E## and ##X_m## are both the (positive) distance traveled by the earth and meteor since the time ##t=0##. and ##M_m## and ##M_E## are the mass of meteor and earth. and ##v## is (positive) initial speed of the meteor towards the earth and ##t## is time. Also, keep in mind that ##X_m## is increasing faster than ##vt##, so this shows that ##X_E## will always be positive.

    So, anyway, from the equation, if ##v## is larger, then ##X_E## will be smaller at the same value of ##X_m## (except when ##t=0##, of course).