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Another E=1/2mv^2 dilemma increasing energy required?

  1. Jul 1, 2010 #1
    It seems to tell us that it takes more and more energy to increase by the same amount of speed.
    I don't actually expend 3 times as much gas accelerating my car from 50 to 100 as I did going from 0 to 50:


    If e=1/2mv2, it tells us how much 'kinetic' energy exists within a moving object -say relative to us.

    In order to obey the Law of Conservation of Energy, the amount of kinetic energy that exists within this moving object must be equal to the amount of energy used to accelerate it to that speed (friction,etc aside).

    Therefore, this formula also suggests how much energy it would take to accelerate an object to a particular velocity. But, it then also implies that the amount of energy required to accelerate an object increases exponentially as the speed increases, very quickly towards infinity. It would take far far more than twice the energy to double the speed, but I don't believe this is what actually happens.

    (in special relativity, when more energy is required to accelerate an object, it is described as the mass increasing with velocity)


    What am I missing here?

    Thanks.
     
    Last edited: Jul 2, 2010
  2. jcsd
  3. Jul 2, 2010 #2

    Andrew Mason

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    I don't know. Perhaps you could tell us why you don't believe it.

    AM
     
  4. Jul 2, 2010 #3
    Well, I don't know. Perhaps you can be more constructive and provide an example of it happening like that instead of being antagonistic.
     
  5. Jul 2, 2010 #4

    rcgldr

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    It shoud take 3 times as much gas, if you measure how much gas is used per second (versus how much gas is used per mile driven).

    Well quadratically (speed^2).

    It is what happens. However what matter is the point of application of the force involved, you don't get to pick just any frame of reference. For example if you use a car as a model, the speed difference bewteen the car and the road the tires push against is the speed that counts.
     
  6. Jul 2, 2010 #5

    Filip Larsen

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    You may want to define a proper system for your car and then investigate how the fundamental physical relationship between energy and force across the system boundary (i.e. "energy equals force times distance") has significance for the kinetic energy of the car.

    Hint: the square dependency on velocity in the equation for kinetic energy is very much related to how you define your system boundary and reference system. A rocket for instance, as it is usually modeled with reaction mass flowing out of its system boundary, is "capable" of a linear increase in velocity for a linear expense of energy.
     
  7. Jul 2, 2010 #6
    Don't get your Tachometer confused with your speed; the way your (internal combustion engine) car works is by having the cylinders turn a crank shaft that then goes through a series of gear to change the rotation speed of the tires. The gears exchange rotational speed for torque (rotational force). So in first gear, you're applying a lot of rotational power to the wheels to get them turning. This extra torque is used to both increase your speed and to overcome rolling resistance. Once you gain some speed, your rolling resistance, the force of which is independent of speed, becomes a small component relative to your car's power and you can start to use gears with less torque and more rotational speed. As your speed increases, you have 2 components to overcome. The first is power required to change in momentum, given by [tex]\frac{d\frac{\frac{1}{2} p^{2}}{m}}{dt}[/tex] and second the drag caused by moving through the air, which is given by the cube of the momentum (velocity). For these speeds, your tachometer may read a low RPM but that's because it's spinning the wheels like crazy to maintain your speed. So it's not that you're using linear power to achieve the same acceleration regardless of initial velocity, it's just that the power is eaten up by different elements beyond simple acceleration before it gets to your car.

    You can get a clearer example of the relationship between speed and Energy [tex]\left(\frac{dW}{dt}\right)[/tex] with an EV, since EVs don't have gear boxes. There, the power is directly transformed into overcoming rolling resistance, overcoming drag, climbing hills (changing the potential energy) and of course accelerating.
     
    Last edited: Jul 2, 2010
  8. Jul 2, 2010 #7
    I think the idea confusing the OP is the relative nature of velocity. Perhaps a simplified rocket example will help.

    Rocket starts a 0 velocity. It eject a given mass at a given velocity to accelerate to a given speed. Now to double the velocity you simply need to eject another equal mass at equal velocity. So the question is why does the second mass ejection more than doubles the kinetic energy relative to the starting point? The second mass ejection also has kinetic energy that the first mass ejection provided it. So the kinetic energy of the two masses that were ejected were not the same from the perspective of the starting point. From the starting point perspective, the second ejected mass is moving slower than the first mass that was ejected. After enough mass ejections, the ejected mass is not even moving in the same direction as the ship that ejected it is, or the direction of the first masses that were ejected, yet remains moving toward the starting point from the ship perspective.

    This is classic Galilean relativity. It doesn't use Einstein's relativity at all.
     
  9. Jul 2, 2010 #8

    Andrew Mason

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    I wasn't being antagonistic. I was asking a simple question. In order for us to determine what you are missing, we have to understand your problem. You appear to be saying that you don't believe that kinetic energy is equal to mv^2/2. How are we supposed to pinpoint the error you are making if you don't explain why you don't accept this? Are we supposed to just guess why you don't accept it? Do you have evidence that it is incorrect?

    You use the example:
    What evidence do you have that this is not true? (ignore the inefficiency of the internal combustion engine as a function of speed and forces of resistance as a function of speed).

    Do you think your car would do only twice as much damage in hitting a brick wall at 100 kph as in in striking it at 50 kph? Do you think that slowing a car from 100 kph to 0 using your mechanical brakes will only generate twice as much heat as slowing from 50 kph to 0?

    AM
     
    Last edited: Jul 2, 2010
  10. Jul 2, 2010 #9
    Clearly - if I "don't think it is true" but it "is" but I am wrong then this is the source of my confusion and the reason for my question. Thanks to all the really smart and very helpful people who were able to see this, they generously cleared up that confusion with meaningful responses.
     
  11. Jul 2, 2010 #10
    It is often quiet difficult intuit from what perspective a question is predicated on. I often get the assumed perspective wrong, and give answers that aren't very useful. Many people don't realize that answers suitable to resolve their questions also often have explanations, requiring a lot more depth to consider. Sometimes we are stuck simply with: that's the rules nature gives us.

    That someone ask for clarification is a commendable act, not an act of antagonism. This perception of being antagonistic is why I just throw out a presumptuous answer and let the cards fall where they may.
     
  12. Jul 2, 2010 #11

    Cleonis

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    I think I can home in on the problem.
    Please bear with me, I'm doing this step by step.


    Let me discuss the following case: a marble is hurled at a chunk of clay. The collision between the marble and the clay is completely inelastic. That is, as the clay deforms all the kinetic energy of the marble is transformed to thermal energy.

    As you point out: the kinetic energy is proportional to the square of the velocity. Twice the velocity gives four times as much thermal energy, triple the velocity gives nine times as much thermal energy, etc.


    I assume now that the determining factor for heat generated is the amount of deformation. I assume that this relation is proportional: penetrating twice as deep gives twice the amount of thermal energy, penetrating four times as deep gives four times as much thermal energy, etc.
    Put differently, the amount of heat generated is proportional to distance traveled in the clay.

    For simplicity I assume that the resistance of the clay against penetration is a constant. So as the marble penetrates the clay it is decelerated by a constant force, giving a uniform deceleration.


    - If the marble travels with a velocity of 10 m/s and the deceleration is 100 m/s^2 then after 0.1 second the marble has come to a halt. The marble will have penetrated 0.05 meter into the clay.

    - If the marble travels with a velocity of 20 m/s (same deceleration: 100 m/s^2) then after 0.2 second the marble has come to a halt. The marble will have penetrated 0.2 meter into the clay.

    So why has the faster marble penetrated so much deeper? The point is, by the time that the marble has been slowed down to 10 m/s the marble has already traveled 0.15 meter through the clay.
    As described by the expression

    [tex] \Delta s = \frac{1}{2} a t^2 [/tex]

    The nature of acceleration and deceleration is that distance traveled is proportional to the square of the amount of time.
    The energy amount is not proportional to the amount of time: it's proportional to distance traveled.


    Incidentally, accelerating your car is consistent with the above. If you accelerate uniformly to, say, 100 km/s then along the section where you go from 0 to 50 km/s you travel far less distance than along the section where you go from 50 to 100 km/s. Kinetic energy is equal to the work being done, and work done is proportional to distance traveled

    The point is: during the evaluation you must not shift your point of reference for distance traveled. If you do shift that point then you end up with self-contradiction.
     
  13. Jul 2, 2010 #12

    Andrew Mason

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    No. Thinking it is not true when it is true is not the source of your confusion. It is the result of it. For some reason, you think it does not take 3 x as much energy to go from 50-100 as from 0-50. I was just asking why you did not think it is true. Apparently, you have difficulty accepting that this is a legitimate question, since you think I am being antagonistic by asking it. Problem-solving in physics, and in any other field, is often about asking the right questions.

    AM
     
    Last edited: Jul 2, 2010
  14. Jul 2, 2010 #13

    russ_watters

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    Yeah actually it does. Check the acceleration specs of any car. It takes MUCH longer to go from 30 to 60 than from 0 to 30 for exactly that reason.
     
  15. Jul 2, 2010 #14

    Cleonis

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    Hi Russ, your comment makes me curious indeed:

    What is your opinion about the following scenario:
    A magnetically levitated train carriage inside a tube that is pumped out to a vacuum. For simplicity assume there is zero friction. Assume the propulsion mechanism gives the carriage uniform acceleration.

    In those circumstances, if it takes x seconds to accelerate from 0 to 500 km/s, how much time will it take to accelerate from 500 to 1000 km/s?

    Cleonis
     
  16. Jul 2, 2010 #15

    Andrew Mason

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    Perhaps you mean m/s. If the force/acceleration is constant, the time is the same: [itex]\Delta t = \Delta v/a[/itex]. The change in momentum going from 0-500 is the same as going from 500-1000 so the impulse [itex]F\Delta t = m\Delta v[/itex] is the same.

    But in order to maintain constant acceleration (force), the rate of energy output (power) will have to increase with speed. dW/dt = d(Fs)/dt = Fds/dt = Fv So the instantaneous power is proportional to speed.

    With a car, the greatest acceleration is in the lowest gears. Acceleration decreases in higher gears, so it usually takes longer to go from 50-100 than from 0-50.

    AM
     
    Last edited: Jul 2, 2010
  17. Jul 3, 2010 #16

    Cleonis

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    Interestingly, this is the content of the question by the thread starter.

    I will stay with the frictionless case. A maglev train, operating in a tube pumped to vacuum. Assume friction is zero
    Assume the power source is batteries onboard the maglev train.

    Scenario:
    The maglev train accelerates in 2 phases: first from 0 units of velocity to 500 units of velocity, then from 500 units of velocity to 1000 units of velocity. In each phase of the acceleration history the same force is generated.

    Question:
    At what rate will the batteries be drained? Will the batteries drain faster during the second phase, accelerating from 500 units to 1000 units?

    I think the thread starter seeks an intuitive answer to the question: given the same force is exerted throughout, why does power consumption increase with velocity?
    In other words, it's a question about the nature of propulsion.
     
  18. Jul 3, 2010 #17

    Andrew Mason

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    Again, I have to ask the question: why would one think it does not? Science is based on evidence, not intuition.

    It may not be intuitive that F = ma . If someone says, well I don't intuitively think F=ma the only question I can ask is: why do you not think F=ma?

    Work/energy is a quantity defined as Force x distance. If a force is applied through a certain distance will generate a certain amount of heat, you double the heat generated if you double the distance.

    If you apply a force through a certain distance to move the train (to a speed of v), and then apply a the same force to the train but in the opposite direction, the train will take that same distance to stop. If you now apply the same force through 4 x the original distance to accelerate the train (thereby reaching a speed of 2v), you will have 4 x the original braking distance. It is just a consequence of Newton's second law.

    AM
     
  19. Jul 3, 2010 #18

    Cleonis

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    Indeed, what you are arguing here was argued by me earlier in this thread in https://www.physicsforums.com/showpost.php?p=2784675&postcount=11".

    The thing is, you are invoking a definition of work in your story, which is putting the cart before the horse. There is a physical reason to define work that way, and to explain one needs to show that reason.

    So far https://www.physicsforums.com/showpost.php?p=2784675&postcount=11" in this thread is my best shot at explanation. I'm not satisfied with it; most likely it can be formulated more concisely.
     
    Last edited by a moderator: Apr 25, 2017
  20. Jul 3, 2010 #19

    Andrew Mason

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    The reason energy is defined that way is that this quantity: energy = ability to do work (work = force x distance); is always conserved in some form. That is shown by experiment. The fact that energy is conserved in all physical interactions is what makes it a useful quantity.

    But that was not the OP's question. He was stating a premise that energy is not conserved (effectively, because he said you don't need to add three times as much energy to go from 50-100 as from 0-50) and was wondering what he was missing. Obviously, the problem is his premise. To answer his question, one has to find out why he adopted such a premise.

    AM
     
    Last edited by a moderator: Apr 25, 2017
  21. Jul 4, 2010 #20

    Cleonis

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    I'm having another stab at answering the OP's question.

    First, let me recapitulate the difference between Energy and Power (as defined in physics language):
    - Energy is a quantity.
    - Power is Energy throughput per unit of time.

    When a weight descends in a controlled manner the amount of gravitatonal potential energy that is released is proportional to how much the weight descends. Example: a pendulum clock with descending weights to sustain its operation.
    Expressed mathematically:

    [tex] E_p = mgh [/tex]

    The weight descends at a uniform velocity, so in this case the Power is uniform.


    Next: I assume that if the weight is allowed to descend in free fall the gain in energy will equally be proportional to how far it has dropped.

    We have:

    [tex] s = \tfrac{1}{2} \cdot a \cdot t^2 [/tex]
    [tex] v = a \cdot t [/tex]

    I replace the distance traveled 's', with 'h' for height, and acceleration 'a' with 'g' for gravitational acceleration.

    [tex] h = \tfrac{1}{2} \cdot g \cdot t^2 [/tex]
    [tex] v = g \cdot t [/tex]

    Using the above expressions I derive an expression without the variable 't':

    [tex] h = \tfrac{1}{2} \cdot g \cdot \left (\frac{v}{g} \right )^2 [/tex]

    Which rearranges to:

    [tex] g \cdot h = \tfrac{1}{2} \cdot v^2 [/tex]

    Multiplying both sides with 'm' gives an expression for kinetic energy as an object comes down in free fall.

    [tex] m \cdot g \cdot h = \tfrac{1}{2} \cdot m \cdot v^2 [/tex]

    As recapitulated: Power is Energy throughput per unit of time.
    In free fall during the first half of the total time 1/4th of the height is traveled, and in the second half of the total time the remaining 3/4th of the height is traveled. The Power is in the same ratio: during free fall the Power delivered by gravity increases quadratically with time.

    This applies for any acceleration.
    So yes, if you have a car that is capable of uniform acceleration from 0 to 60, then the time it takes to accelerate from 30 to 60 is the same as the time it took to accelerate from 0 to 30. But the engine must deliver three times as much power during the push from 30 to 60.

    But hang on, haven't we been spoonfed the principle of relativity of inertial motion, according to which accelerating from 30 to 60 is the same as accelerating from 0 to 30? What gives?

    The thing is, to increase the velocity of the car the wheels must grip the road. To accelerate you must have leverage in one form or another. Once the car is up to a velocity of 30 then the thing that the wheels must grip is a surface that is moving relative to the car at a velocity of 30. That makes it harder to get the necessary leverage.
     
  22. Jul 4, 2010 #21

    Andrew Mason

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    One can get the "necessary leverage" using gears. The problem is that it takes 3 times as much average power to accelerate from 0-30 as it does from 30-60 if you are maintaining the same acceleration. So unless the engine can deliver the extra power, the acceleration will fall off as speed increases.

    It is simply a consequence of energy/work being Force x distance. You could ask why is energy/work defined as Force x distance, but that is a different question.

    AM
     
  23. Jul 5, 2010 #22
    If i choose to seperately handle the energy consumsion when there is an increase of velocity or and increase of momentum, am i correct?

    I am thinking if an object gain velocity but giving up mass, it can have the same momentum while different kinetic energy through the trip.
    In this example, gaing velocity alone requires energy.

    On the other hand, if i try to increase the mass of a moving object while maintaining a constant speed, there is a change in momentum but not velocity, howerver, the kinetic energy increase again.
    In this example, gaining moment alone again requires energy.

    So there is two catogory to concern when trying to speed up an object (constant mass), the momentum and the velocity both increase, and both require energy, thats why doubling the speed will more than doubling the kinetic energy, because other than speed, the momentum also double.

    Is that concept correct? Cause even me doubt it. :confused:

    p.s. if my bad english confuse you, do reply and tell me....
     
  24. Jul 5, 2010 #23
    Drop a ball from the 2nd floor window and see how fast it hits the ground. Now go to the 3rd floor (twice as high) and drop it again. Despite the fact that the ball started out with 2x as much potential energy, it will NOT be going 2x as fast when it hits the ground, even disregarding air resistance.

    The reason for this is that while gravity exerts the same force on the ball throughout its fall, the ball is going much slower when it first starts falling...therefore gravity acts on it for a longer time. Once it passes the half way point, it's already moving pretty quick and gravity doesn't have that much time to accelerate it further.

    So, the ball will hit the ground only about 1.42x as fast. In fact, if you want the ball to hit the ground going 2x as fast, you will indeed have to go at least 4x higher....to the 5th floor.

    Do the math, and see. If you don't believer the math, find a ball and a building and check it yourself. It might not be intuitive at first, but that's the way it is. You will find that it works the same everywhere you go, whether with your car, your bike, a ball, or a rocket (althouth a rocket has some complicating factors).
     
  25. Jul 5, 2010 #24

    Cleonis

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    AM, you have a tendency to stop just when things get interesting.


    Work = force * distance

    There is the example of the experiments by James Prescott Joule. He set up weights to go down in a controlled descent, driving paddles, churning water in a very well insulated cannister. He found that the temperature rise due to the churning was proportional to the distance traveled by the weights.

    No doubt numerous other experiments have been conducted to establish that with a controlled release of potential energy (uniform velocity) the work done is proportional to distance traveled.

    Acceleration
    In my opinion it's far from trivial that in the case of acceleration the work done is also proportional to the distance traveled. I find that quite marvelous.

    In https://www.physicsforums.com/showpost.php?p=2786672&postcount=20" of this thread I show that if you demand self-consistency then it is implied that during acceleration work done must be proportional to distance traveled.

    My gut feeling says there is room for deeper explanation in this case, but I can't put my finger on it.
     
    Last edited by a moderator: Apr 25, 2017
  26. Jul 5, 2010 #25

    russ_watters

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    Cleonis your statement about the principle of relativity implies you don't know how it works/ what it means. Could you expand on that part so we can pinpoint your misunderstanding?
     
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