Messed up classical mechanics problem:

[Curl]

This is really simple but I can't figure it out. I was on a bus when I thought of this:

Say I'm sitting in the back of a bus which is traveling on a flat surface, and accelerating with a constant acceleration (forward). Now I get up from my seat in the back and make my way to the front of the bus and take a seat at the front. Shortly after I sit down, the bus stops accelerating and is cruising at constant speed.
When I was walking (say at constant rate relative to the floor of the bus) I felt like I was doing work against a "fictitious" gravity force, so it felt like I gained some "potential" energy. Now that the bus is not accelerating anymore, it seems like that potential is gone and some energy is lost. Where did it go?

 

[Drakkith]

Hmm. To throw out a guess I'd say that your potential energy was transferred to the bus. When you got up and started to move you slowed the acceleration of the bus. When you sat back down you momentarily increased the acceleration. But I'm really not sure. It's kind of confusing considering the bus has to expend energy to accelerate and all that.

 

[boneh3ad]

If this situation occurred, when the bus reached the constant velocity you would feel like you were about to fly forward since you were then applying more force to walk than you needed to maintain a constant speed with respect to the bus. Either you do suddenly move forward, in which case the potential energy you speak of is now kinetic energy. Otherwise, you would have to brace yourself a bit to lower your own "walking force" in which case that potential energy would be dissipated by your muscles as heat in the process of bracing yourself.

 

[gsal]

when you were first seating, the back of the seat was doing all the work of pushing your body mass to make it keep up with the acceleration of the bus...

...when you got up, your own legs and your skillful balancing act had to do the work of making your body keep up with the acceleration of the bus...and even move at a constant speed respect to it...so, all the work of yours was not in vane...you have kept up with the bus and also gained such acceleration (and obtained the same final speed)...that's why you are still a v=0 respect to the bus and can find a seat to sit on...otherwise, you would be out the back window...

...so, did you gain some "potential" (kinetic rahter) energy...sure, you are going at the final speed achieved...and now that the bus is not accelerating either, nothing is felt...everybody is going at the same final speed.

 

[Curl]

"Otherwise, you would have to brace yourself a bit to lower your own "walking force" in which case that potential energy would be dissipated by your muscles as heat in the process of bracing yourself."
Do you not know how to read?

"Shortly after I sit down, the bus stops accelerating and is cruising at constant speed."

I'm not walking when the bus stops accelerating.

--
"To throw out a guess I'd say that your potential energy was transferred to the bus. When you got up and started to move you slowed the acceleration of the bus. When you sat back down you momentarily increased the acceleration."

This is does not answer anything. It has to do with the symmetry of the walk, however a very easy way to see it is that I can vary the length of my walk arbitrarily while my start/stop motion is always the same.

--
gsal, I'm not following what you said. The engine doesn't feel anything different when you are walking, however your legs are doing work. This work goes... where?

 

[gsal]

Curl:

Actually, the engine feels something when you are walking...ever so slightly, but your motion is moving the center of mass of bus and passengers a little more forward and "accelerating" more than if you had stayed in your seat...

In any case, the work that you are doing goes towards making your own mass keep up with the acceleration of the bus...even you do not walk and just stand up...you need to take stand so that you do not go out the back window...so, your work did not go to waste...you were able to keep up with the motion of the bus...so, you are right, you gain something, but not potential energy, you gained kinetic energy and are now going at the same speed as the bus.

The only reason why you do not feel this "potential" energy after you sit down and the bus is now at constant speed is precisely because it has reached its final speed and is not accelerating you anymore (F=ma=0) and so you feel no force.

Does this help?

 

[Drakkith]

This is does not answer anything. It has to do with the symmetry of the walk, however a very easy way to see it is that I can vary the length of my walk arbitrarily while my start/stop motion is always the same.

As I said, I don't know. What kind of potential energy do you think you have gained?

gsal, I'm not following what you said. The engine doesn't feel anything different when you are walking, however your legs are doing work. This work goes... where?

Sure it does. You are applying a force against the bus, the engine must compensate for this tiny extra force against it if it wants to keep accelerating at the same rate or it must accelerate slower.

 

[willem2]

As I said, I don't know. What kind of potential energy do you think you have gained?

Actually, it makes sense to talk about potential energy in reference frame which is constantly accelerated. As we can see from this problem, it does not make sense to do so, if the acceleration is not constant.

Sure it does. You are applying a force against the bus, the engine must compensate for this tiny extra force against it if it wants to keep accelerating at the same rate or it must accelerate slower.

Actually the result of the entire movement, is that you do some work to accelerate yourself, which does not have to be done by the engine of the bus.

When you get up and accelerate yourself to a constant forward speed, you will indeed slow down the bus somewhat, but when you sit down again, and lose your speed with respect to the bus, you will give this momentum back to the bus, and since the bus is now moving faster, it will gain more energy than it lost when you got up.
When you walk forward with constant speed, this will have no effect on the bus, and the work that you do will go into increasing your own kinetic energy. The power needed for you to accelerate is Force * (speed of bus + walking speed), so this is larger than when you're not moving with respect to the bus.

 

[Drakkith]

Actually the result of the entire movement, is that you do some work to accelerate yourself, which does not have to be done by the engine of the bus.

You accelerate yourself by pushing against the bus opposite to the direction of acceleration. How does the engine not have to do work for that?
Edit: Note that if the engine does no extra work then the rate of acceleration slows. So the engine either does more work to keep the same acceleration, or it must accelerate slower.

When you get up and accelerate yourself to a constant forward speed, you will indeed slow down the bus somewhat, but when you sit down again, and lose your speed with respect to the bus, you will give this momentum back to the bus, and since the bus is now moving faster, it will gain more energy than it lost when you got up.

Remember that the bus is accelerating the whole time you are moving, it is not traveling at a constant velocity. This means that you must continually work to keep a constant velocity relative to the bus. When you stop walking and go to sit down the bus gains nothing, as you haven't done anything to give momentum back to it. Each time you push off with one of your feet to move forward you apply a force against the bus. When that step has ended the bus simply regains its initial acceleration unless you take another step to keep the same velocity.

When you walk forward with constant speed, this will have no effect on the bus, and the work that you do will go into increasing your own kinetic energy. The power needed for you to accelerate is Force * (speed of bus + walking speed), so this is larger than when you're not moving with respect to the bus.

I don't think so. Imagine you are in space. Pushing off the side of your ship will mean that it moves slightly in the opposite direction that you do. When you reach your destination and push against a wall or whatever to stop yourself both you and the ship will cease moving. There is nothing about an accelerating frame that changes this.

My conclusion here is that the potential energy gained by moving against the bus is NOT lost when the bus ceases accelerating. Once the bus stops accelerating get up and walk back to your original seat. The energy you accumulated will be given back to the bus, accelerating it slightly.

 

[willem2]

You accelerate yourself by pushing against the bus opposite to the direction of acceleration. How does the engine not have to do work for that?

As I said, you push more when you get moving, you push the same amount when you move with constant speed, and you push less when you stop. The average force you exert backwards is obviously the same (because the end velocity is the same, wether you move or remain in your seat), but if you move forwards, more of this force is excerted when the bus moves slower.

This means that the bus will have to spend less energy to accelerate you, because it
takes less power to push a slow moving object than a fast moving object with the same force.

My conclusion here is that the potential energy gained by moving against the bus is NOT lost when the bus ceases accelerating. Once the bus stops accelerating get up and walk back to your original seat. The energy you accumulated will be given back to the bus, accelerating it slightly.

Moving back to your place after the bus has stopped doesn't take any energy in theory,
because it can be done without friction and arbitrarily slow. So according to you, the energy that was spend moving forwards simply disappears.

 

[boneh3ad]

"Otherwise, you would have to brace yourself a bit to lower your own "walking force" in which case that potential energy would be dissipated by your muscles as heat in the process of bracing yourself."
Do you not know how to read?

I feel the need to point out that if you feel the need to act like an ******* you will have a hard time getting anywhere in life.

 

[zoobyshoe]

In any case, the work that you are doing goes towards making your own mass keep up with the acceleration of the bus...even you do not walk and just stand up...you need to take stand so that you do not go out the back window...so, your work did not go to waste...you were able to keep up with the motion of the bus...so, you are right, you gain something, but not potential energy, you gained kinetic energy and are now going at the same speed as the bus.

I agree. If he had remained seated, force from the back of his seat would have accelerated him with little effort on his part. Since he decides to stand up, now he must expend extra effort by holding himself at an unusual forward angle and pushing harder with his legs. He's reduced the surface area of his contact with the bus by shifting it from his back to his feet. He now has to employ all kinds of muscles that wouldn't be involved if he'd stayed seated. If he didn't make extra effort to stay attached to the bus now that he's standing, the bus would accelerate forward without him (at least, he'd fall on his back). He gains kinetic energy, but he does it the hard way.

On top of it, he walks forward while the bus is still accelerating. He accelerates himself, with each step, against an already accelerating bus, much like walking up an escalator that is already moving up, both (his and the escalator's efforts) against the acceleration of gravity. That is why it "feels" like he's gaining PE: he would be if he were on an escalator, but in the case of the bus the acceleration is horizontal, not vertical. His gain for all the effort of walking forward is that he'll arrive at his destination slightly earlier. He's shortened his trip by a little less than a bus length.

 

[AlephZero]

Calling this "potential energy" makes no sense to me. Working in an inertial reference frame fixed to the road, you gained more kinetic energy by walking forwards that you would have done by staying in your seat. When you sat down, the "extra" KE was lost in the work you did to slow yourself down to the same speed as the bus.

What else is there to understand here? Debates about the effect on the acceleration of the bus, the work done by the bus engine, etc are hypothetical unless you carefully and completely define two different scenarios and compare them with each other.

 

[Curl]

The paradox was that I could have waited for the bus to stop accelerating and then change seats -> then it would not require any net work from my part.
If I change seats while the bus is accelerating, it seems like I put in work then it disappeared.
willem2 got the correct answer: I give energy to the bus since I slow down at a point in time when the bus is traveling faster than when I began walking.

 

[boneh3ad]

It wouldn't require net work on your part but it would require work by the seat to decelerate you with the bus.

If you walk forward, of course you did work. It didn't disappear. You end up moving a total of however many rows farther ahead than you would had you remained seated, which means you had to do work to move that extra distance.

 

[Drakkith]

As I said, you push more when you get moving, you push the same amount when you move with constant speed, and you push less when you stop. The average force you exert backwards is obviously the same (because the end velocity is the same, wether you move or remain in your seat), but if you move forwards, more of this force is excerted when the bus moves slower.

This means that the bus will have to spend less energy to accelerate you, because it
takes less power to push a slow moving object than a fast moving object with the same force.

I'm sorry I can't even follow this, it doesn't seem to make any sense. I'm not saying it's incorrect, I simply can't follow it the way it's written.

Moving back to your place after the bus has stopped doesn't take any energy in theory,
because it can be done without friction and arbitrarily slow. So according to you, the energy that was spend moving forwards simply disappears.

Stop pushing and let yourself "fall" to the back. Where in this does any energy disappear? YOU spent energy to do work to move yourself forward. Where did it go? It went to counteract the acceleration of the bus. You can move back to the rear of the bus by spending only a little energy to jump up a bit or something exactly the same way a cliff climber can get back to Earth by pushing off the side and letting go and falling down.

The paradox was that I could have waited for the bus to stop accelerating and then change seats -> then it would not require any net work from my part.
If I change seats while the bus is accelerating, it seems like I put in work then it disappeared.
willem2 got the correct answer: I give energy to the bus since I slow down at a point in time when the bus is traveling faster than when I began walking.

It required work on your part to move forward. The work goes into counteracting the bus's acceleration while you move forward.

 

[chrisbaird]

The problem is that you are switching frames when the bus stops accelerating, from an accelerating frame with a fictitious force, to an inertial frame. Energy is only conserved in the same frame (E_before in frame 1 only equals E_after in frame 1). Switching frames half-way through without transforming all variables properly will always lead to trouble (like disappearing energy.) Although not as intuitive, it's probably best to frame the problem relative to the ground the entire time.

 

[willem2]

It required work on your part to move forward. The work goes into counteracting the bus's acceleration while you move forward.

The problem with this, that if you put in work, it should increase the kinetic energy of the bus, so the work has to go in helping the bus accelerate.

It's even clearer if you do the reverse move.
Put something like a cart at the front of the bus. If you release it, it will start to roll backwards. Now if you apply the brakes to bring the cart to a stop relative to the bus, heat will be produced. The only source of this heat can be the engine of the bus.

 

[Drakkith]

The problem with this, that if you put in work, it should increase the kinetic energy of the bus, so the work has to go in helping the bus accelerate.

I can't see how you are coming to this conclusion. You are applying a force opposite to another force, with the net force being the difference in the two. If you reduce the force applied to a mass you also reduce it's acceleration.

It's even clearer if you do the reverse move.
Put something like a cart at the front of the bus. If you release it, it will start to roll backwards. Now if you apply the brakes to bring the cart to a stop relative to the bus, heat will be produced. The only source of this heat can be the engine of the bus.

A clear example of trading potential energy (for this accelerating frame) for kinetic and then thermal energy. While the cart was rolling the bus was accelerating faster because it had less mass to accelerate. When the brakes were engaged the cart came to a stop, causing the bus to revert to it's previous rate of acceleration. Some of the energy was transferred to heat via the brakes and is now unrecoverable.

 

[willem2]

I can't see how you are coming to this conclusion. You are applying a force opposite to another force, with the net force being the difference in the two. If you reduce the force applied to a mass you also reduce it's acceleration.

Let's try some numbers.

at t=0 the bus is traveling at 10 m/s, and for t>0 it has a constant acceleration of 1m/s^2.

Now let's compare two passengers, both with a mass of 80kg.

The first one stays in his seat for the next 12 seconds. The force that the bus excerts on this person is a constant 80 N, and the engine of the bus will have to exert an extra 80 N on the ground to compensate. The bus moves 12*(average speed) = 12*(10+22)/2 = 192 m during this time, so the extra power needed to accelerate the passenger is 80*192 = 15360 J.

Now we have another passenger, who accelerates with 1m/s^2 with respect to the bus during 1 second, then moves to the front of the bus with constant speed for 10 seconds, and then decelerates with 1m/s for 1 second, so he has the same speed of the bus after 12 seconds.
During the first second the total acceleration is 2m/s^2, so force the bus has to exert on him is 160 N. The average speed of the bus during this interval is 10.5 m/s, the distance that the bus moves along the road is 10.5 m, and the extra energy needed for acceleration during this interval is 10.5 * 160 = 1680 J.

during the 10 seconds of constant speed with respect to the bus, the passenger is accelerated by the normal 1m/s^2, so the force that the bus excerts is 80 N. The energy that the bus needs to do this is 80*(average speed) * time = 80 * (11+21)/2 * 10 = 12800J.

During the last second, the acceleration of the passenger is 0 m/s, so no force is needed and the bus doesn't have to use any energy to accelerate the passenger during this time.

If we add up the energy transferred by the bus to the passenger during the entire 12 seconds, we get 1680+12800+0 = 14480 J.

The difference: 15360-14480 = 880 J, is the energy that the passenger put in, who pushed the bus 0.5m backwards with a force of 160 N in the first second, and then pushed it 10m backwards with a force of 80 N. during the next 10 seconds. (and did no pushing in the last second) 160*0.5 + 80*10 = 880N.

 

[willem2]

The bus would have less acceleration during the 1st second because the person is pushing against it.

I'm assuming constant acceleration of the bus, and computing the energy needed to accelerate the passenger. I don't think it will matter much since the bus will be about 100x as heavy as the passenger, so would still be accelerating with 0.99 m/s or so during the first second, and then would accelerate at 1.01 m/s during the last second, ending up with the same speed.

 

[Drakkith]

I'm assuming constant acceleration of the bus, and computing the energy needed to accelerate the passenger. I don't think it will matter much since the bus will be about 100x as heavy as the passenger, so would still be accelerating with 0.99 m/s or so during the first second, and then would accelerate at 1.01 m/s during the last second, ending up with the same speed.

Yeah I figured that as soon as I posted. I didn't think it was up long enough for anyone to quote. I'm going over the math right now.

 

[Drakkith]

I'm sorry willem, I still don't see why your saying the person helps accelerate the bus.

 

[Curl]

It is really simple, don't overthink it, just go back and read his first post, you'll see it instantly.

 

[willem2]

I'm sorry willem, I still don't see why your saying the person helps accelerate the bus.

I'm saying that because the bus expends 880 J less than if the passenger remained seated. Of course the passenger would help the bus accelerate much more by not being in it, since the bus engine still has to produce 14480 J to accelerate the passenger.

 

[Drakkith]

I'm saying that because the bus expends 880 J less than if the passenger remained seated. Of course the passenger would help the bus accelerate much more by not being in it, since the bus engine still has to produce 14480 J to accelerate the passenger.

Is your math correct? Shouldn't you be including the movement of the person on the bus? The person moves somewhere around 10 meters further than the bus does.

 

[willem2]

Is your math correct? Shouldn't you be including the movement of the person on the bus? The person moves somewhere around 10 meters further than the bus does.

Note that the energy the bus needs to accelerate the passenger is the integral of (speed of the bus) * (force excerted on the passenger) over the time interval of 12 seconds, and the position or the speed of the passenger doesn't come into this.
In the earthbound frame the passenger doesn't get extra potential energy by walking to the front of the bus.

 

[Drakkith]

Note that the energy the bus needs to accelerate the passenger is the integral of (speed of the bus) * (force excerted on the passenger) over the time interval of 12 seconds, and the position or the speed of the passenger doesn't come into this.
In the earthbound frame the passenger doesn't get extra potential energy by walking to the front of the bus.

I'm sorry, I simply don't believe your math is correct. There is no way the bus spends LESS energy on the moving person.
The only difference is 80 N over 1 second is taken from the end and placed in the beginning. At minimum the two should equal out.

 

[Drakkith]

Ok all, I've spent the better part of 3-4 hours today taking a self-taught crash course in basic physics to try to understand this. (Alot was trying to calculate the energy, but I can't do it tonight.) Does the following make sense?

Moving forward against the bus required work on the passengers part. Take the Cart at the front of the bus example that willem gave. If we move the cart to the front of the bus from the back we give it potential energy in the accelerating frame. This is clear because releasing the cart would convert the potential energy into kinetic energy as the cart accelerates down the middle and slams into the back of the bus.

ALSO, from the frame of the ground, the cart initially gained kinetic energy faster than the bus did, as it had to accelerate more than the bus to gain speed relative to the bus. This is counteracted when the cart reaches the front and decelerates. From the ground's point of view it wasn't accelerating at all, but "coasting" along, so its earlier gain is equaled out by no gain now. So at the end it has equal velocity and acceleration to the bus. Note that at no time was kinetic energy "lost". From the frame of you on the bus it was converted into potential energy after you stopped applying a force. From the ground frame you simply stopped gaining kinetic energy for a second while the bus continued to gain KE for that second.

Let's assume that the mass of the bus is 10,000 kg including passengers and cargo.
The force required to accelerate the bus at 1 m/s^2 is: F=1*10,000
So it requires 10,000 Newtons to accelerate the bus.

When the passenger is accelerated at 1 m/s^2 relative to the bus, they must apply a force of 80 N against the bus, opposite of it's acceleration. So net force for the bus is now 9,920 N, meaning an acceleration of 0.992 m/s^2 UNLESS the engine uses more fuel to apply the lost 0.008 N of force. So, assuming the engine does NOT compensate, then the acceleration on the individual as seen from the ground is 1.992 m/s^2.

 

[willem2]

Let's assume that the mass of the bus is 10,000 kg including passengers and cargo.
The force required to accelerate the bus at 1 m/s^2 is: F=1*10,000
So it requires 10,000 Newtons to accelerate the bus.

When the passenger is accelerated at 1 m/s^2 relative to the bus, they must apply a force of 80 N against the bus, opposite of it's acceleration. So net force for the bus is now 9,920 N, meaning an acceleration of 0.992 m/s^2 UNLESS the engine uses more fuel to apply the lost 0.008 N of force. So, assuming the engine does NOT compensate, then the acceleration on the individual as seen from the ground is 1.992 m/s^2.

This looks ok. The total force needed to accelerate the passenger is m.a = 80 * 1.992m/s = 159.36 N.

The passenger moves 0.5 m relative to the bus, so needs to input 159.36 * 0.5 = 79.68 J of energy to accelerate himself.

The bus moves v_0 * t + (1/2) a t^2 = 10 *1 + (1/2) * 0.992 * 1^2 = 10.496 meters along the road, so it needs to spend 10.496 * 159.36 = 1672.6 J to accelerate the passenger.
Of course the bus would have needed 10.5 * 80 = 840 J to accelerate a passenger who doesn't move during the first second.

 

[daqddyo1]

In Curl's initial statement, there is no change in potenial energy (gravitational or otherwise). The work done by Curl while walking to the front of the bus while it is accelerating is increasing his kinetic energy.
If he were sitting in his seat during the acceleration (the easier way), the force exerted by his seat back acting through the distance the bus accelerated would do the work necessary to increase Curl's kinetic energy.

 

[zoobyshoe]

This is really simple but I can't figure it out. I was on a bus when I thought of this:

Say I'm sitting in the back of a bus which is traveling on a flat surface, and accelerating with a constant acceleration (forward). Now I get up from my seat in the back and make my way to the front of the bus and take a seat at the front. Shortly after I sit down, the bus stops accelerating and is cruising at constant speed.
When I was walking (say at constant rate relative to the floor of the bus) I felt like I was doing work against a "fictitious" gravity force, so it felt like I gained some "potential" energy. Now that the bus is not accelerating anymore, it seems like that potential is gone and some energy is lost. Where did it go?

Curl, how fast do you estimate the bus' instantaneous velocity was when you started? How much was it when you sat down again, and how many seconds elapsed in between?