Physic: WORK

[Gpan]

Ok we all know that W = Fdcos0. So if the angle is 90, then there is no work perform. However, I keep thinking about this scenario.

Let's say there's a toy car with initial V= 1 m/s going in one direction. Now if you put your hand on top of the car and slightly press downward. This action will increase the weight of this car and will change the velocity of the car in a way that it will slow down. The question is did you do any work? The angle is 90 at this point, so is the W=0? I don't see it.

Because surely you have contributed something into the system, a downward force F. And that force F must have done some work to slow down the car, right?

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[iA-MD2013]

Ok we all know that W = Fdcos0. So if the angle is 90, then there is no work perform. However, I keep thinking about this scenario.

Let's say there's a toy car with initial V= 1 m/s going in one direction. Now if you put your hand on top of the car and slightly press downward. This action will increase the weight of this car and will change the velocity of the car in a way that it will slow down. The question is did you do any work? The angle is 90 at this point, so is the W=0? I don't see it.

Because surely you have contributed something into the system, a downward force F. And that force F must have done some work to slow down the car, right?

good question. so the way i see it is this...
you apply a force in the vertical direc, but what is the resut of the force? well, youve increased the normal force. increasing normal force will increase the frictional force. now the frictional force points in the horizontal direc, against the direc of motion, causing the deceleration of the car. SO, your force does do work because of its change in frictional force. negative work.
is this a valid explanation? any other opinions?

 

[blastula]

applying force at 90 deg is torque. before you applied this force the car was in rotational equilibrium. the car slows down to counter the torque applied

torque = f x d sin

 

[RSAgator]

well I think that the scenario you're thinking of could be more accurately depicted by momentum with a completely inelastic collision (which the way i view it is essentially what's happening). Lets say you apply a force of 100N to a 2kg car moving 1m/s. This is the equivalent of applying a mass of 10kg.

m1v1 +m2v2 = (m1+m2)v3

(2)(1) + (10)(0) = (12)v3
v3 = 2/12 = 1/6

thus your velocity has dropped due to conservation of momentum.

 

[iA-MD2013]

well I think that the scenario you're thinking of could be more accurately depicted by momentum with a completely inelastic collision (which the way i view it is essentially what's happening). Lets say you apply a force of 100N to a 2kg car moving 1m/s. This is the equivalent of applying a mass of 10kg.

m1v1 +m2v2 = (m1+m2)v3

(2)(1) + (10)(0) = (12)v3
v3 = 2/12 = 1/6

thus your velocity has dropped due to conservation of momentum.

sure, but have you done work?

 

[RSAgator]

sure, but have you done work?

Ah I didn't even see you'd responded to the op's post.

I am inclined to disagree with you. Think of work as energy. When you do work on something you increase its energy. Now in this case we have two types of energy, heat energy lost due to friction and kinetic energy. While you are increasing the amount of energy that is lost as heat energy, you are not increasing the energy of the system. That extra heat energy that is lost will come from the kinetic energy. Thus the energy in the system is being transferred, but you are not doing work (contributing to the energy of the system). If you were to idealize the situation and assume no heat loss, you would find that the kinetic energy before equals the kinetic energy after, but the velocity changes because effectively the mass changes.

Another reason you can't be right is that hypothetically the frictional coefficient could be 0. If that's the case your answer would imply that if there is no friction there is no work done which doesn't seem like a reasonable assumption. It's often best to neglect friction when considering hypothetical scenarios =)

 

[iA-MD2013]

Ah I didn't even see you'd responded to the op's post.

I am inclined to disagree with you. Think of work as energy. When you do work on something you increase its energy. Now in this case we have two types of energy, heat energy lost due to friction and kinetic energy. While you are increasing the amount of energy that is lost as heat energy, you are not increasing the energy of the system. That extra heat energy that is lost will come from the kinetic energy. Thus the energy in the system is being transferred, but you are not doing work (contributing to the energy of the system). If you were to idealize the situation and assume no heat loss, you would find that the kinetic energy before equals the kinetic energy after, but the velocity changes because effectively the mass changes.

Another reason you can't be right is that hypothetically the frictional coefficient could be 0. If that's the case your answer would imply that if there is no friction there is no work done which doesn't seem like a reasonable assumption. It's often best to neglect friction when considering hypothetical scenarios =)

omg, youre disagreeing with me no wait, wrong facial expression....this one is better:
haha ok, im a really slow typer, so im not going to respond to this until i ask a gsi tomorrow. but, dcohen is really smart, so he is probably right.

 

[BlackSails]

In this example, work is done, but by friction, not by the force pushing down.

 

[Gpan]

In this example, work is done, but by friction, not by the force pushing down.

But without the force pushing down, there would be no work done by friction.

 

[Gpan]

Ah I didn't even see you'd responded to the op's post.

I am inclined to disagree with you. Think of work as energy. When you do work on something you increase its energy. Now in this case we have two types of energy, heat energy lost due to friction and kinetic energy. While you are increasing the amount of energy that is lost as heat energy, you are not increasing the energy of the system. That extra heat energy that is lost will come from the kinetic energy. Thus the energy in the system is being transferred, but you are not doing work (contributing to the energy of the system). If you were to idealize the situation and assume no heat loss, you would find that the kinetic energy before equals the kinetic energy after, but the velocity changes because effectively the mass changes.

Another reason you can't be right is that hypothetically the frictional coefficient could be 0. If that's the case your answer would imply that if there is no friction there is no work done which doesn't seem like a reasonable assumption. It's often best to neglect friction when considering hypothetical scenarios =)

I understand what you were saying, but I don't agree with that bold part. Work does not always have to increase the energy of the system.

Let's say there's a block going downhill with some speed. Its energy right now is the KE. Let's say you apply a force upward parallel to the plane of the hill to stop the block dead on its track. You've obviously done work, and the energy of the block did not increase.

 

[tncekm]

But without the force pushing down, there would be no work done by friction.

If there were no friction, there would have been no work. You just altered the normal force--its no different than saying that putting the toy car on a bigger planet would have created work by increased gravity, regardless of the fact that the force isn't changing the position of the car with respect to the direction of the gravitational field.

 

[RSAgator]

I don't think you're correct. You've converted kinetic energy to heat energy but you haven't done work on the object. Lets say you have a ball moving on a horizontal surface, it clearly has kinetic energy. You apply a force to the ball and the ball stops. Have you added negative energy to the ball, causing it to stop? No, you have converted all of its kinetic energy into heat energy.

Consider friction which is essentially a force opposing your moving object. You have a ball rolling on a surface with friction. Your ball slows down over time and eventually it stops. All of its kinetic energy has been converted to heat energy. If you were to sum up all the heat that was lost as the ball slowed down, it would be equal to the kinetic energy the ball had inititally. Even though there was a "force" (in this case frictional force) being applied to the ball, work wasn't being done on the ball (which would increase total energy) nor was work being done by the ball (which would decrease its total energy).

As for my bolded statement, you can think of many scenarios where essentially you are applying a force to an object and you'll see that energy is conserved and no extra work is done. In this case it's simply that the useful energy (kinetic energy) is converted into useless energy (heat energy) which you can't use again. The scenario is also no different to a spring. A ball with K.E rolls into a spring. The spring applies a force to the ball slowing it down and converting the K.E into potential energy. You have applied a force to the ball but the energy in both the ball and the spring is still the same as the energy in just the ball, only the energy is in different forms. Does that make any more sense?

 

[Gpan]

yup, thanks again.

 

[BlackSails]

But without the force pushing down, there would be no work done by friction.

You place a block on top of this car. Have you done any work? No.(assume the car is flat)

 

[majik1213]

Ok we all know that W = Fdcos0. So if the angle is 90, then there is no work perform. However, I keep thinking about this scenario.

Let's say there's a toy car with initial V= 1 m/s going in one direction. Now if you put your hand on top of the car and slightly press downward. This action will increase the weight of this car and will change the velocity of the car in a way that it will slow down. The question is did you do any work? The angle is 90 at this point, so is the W=0? I don't see it.

Because surely you have contributed something into the system, a downward force F. And that force F must have done some work to slow down the car, right?

work is the integral of a non-variable force times it's displacement .. in its truest sense, work is defined as the change in energy with respect to a change in an external parameter

you're also adding friction to your scenario .. wheels have fricitions, which is proportional to the normal force you exert

 

[Rofeh20]

I don't think you're correct. You've converted kinetic energy to heat energy but you haven't done work on the object. Lets say you have a ball moving on a horizontal surface, it clearly has kinetic energy. You apply a force to the ball and the ball stops. Have you added negative energy to the ball, causing it to stop? No, you have converted all of its kinetic energy into heat energy.

Consider friction which is essentially a force opposing your moving object. You have a ball rolling on a surface with friction. Your ball slows down over time and eventually it stops. All of its kinetic energy has been converted to heat energy. If you were to sum up all the heat that was lost as the ball slowed down, it would be equal to the kinetic energy the ball had inititally. Even though there was a "force" (in this case frictional force) being applied to the ball, work wasn't being done on the ball (which would increase total energy) nor was work being done by the ball (which would decrease its total energy).

Actually, that's not correct. Via the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy.

If an object initially has 100J of KE and then it is stopped it has lost 100J of kinetic energy. Whether that 100J of KE has been converted over to potential energy (conservative conditions) or heat (non-conservative) is irrelevant -- the bottom line is that the object did lose 100J of KE. Thus, the net work done on the object is negative. Equivalently, the net work done by the object is positive.

Remember that the Wby object = -(Won on object)

What you are probably confusing is the concept that energy cannot be created or destroyed. Energy is merely transferred from one form to another (e.g. kinetic into potential, or kinetic into heat). Work is merely the term used to describe the magnitude of the energy transferred.


Edit: Allow me to elaborate further on what was incorrect in your post. The statement that "If you were to sum up all the heat that was lost as the ball slowed down, it would be equal to the kinetic energy the ball had inititally" is correct. The problem came with your next statement: "work wasn't being done on the ball (which would increase total energy) nor was work being done by the ball (which would decrease its total energy)". The frictional force does negative work on the ball, so it does decrease the ball's kinetic energy (and its total mechanical energy). An equal statement would be to say that the ball does positive work on its surroundings.

 

[RSAgator]

Edit: Allow me to elaborate further on what was incorrect in your post. The statement that "[/SIZE][/SIZE][/SIZE][/SIZE]If you were to sum up all the heat that was lost as the ball slowed down, it would be equal to the kinetic energy the ball had inititally" is correct. The problem came with your next statement: "work wasn't being done on the ball (which would increase total energy) nor was work being done by the ball (which would decrease its total energy)". The frictional force does negative work on the ball, so it does decrease the ball's kinetic energy (and its total mechanical energy). An equal statement would be to say that the ball does positive work on its surroundings.

you're quite right, friction is considered negative work. I think the concept of "negative work" is confusing in and of itself and it's better to consider the ball doing work on its surroundings than something doing negative work on the ball. To say "negative work is done on the ball" creates an exception that doesn't need to exist.

Quite simply if an object does work it loses energy and if an object has work done on it it gains energy (I know this isn't a point you're arguing but it is a point I'd like to make). Just as you pointed out, you could easily say an object has negative work done on it but I think it's conceptually wrong to think that this is what is happening. Obviously that's my opinion which counts for very little =).

Thanks for the clarification though. To take it a step further, any time a system loses energy to heat is the system doing work on its surroundings?

 

[BloodySurgeon]

Actually, that's not correct. Via the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy.

If an object initially has 100J of KE and then it is stopped it has lost 100J of kinetic energy. Whether that 100J of KE has been converted over to potential energy (conservative conditions) or heat (non-conservative) is irrelevant -- the bottom line is that the object did lose 100J of KE. Thus, the net work done on the object is negative. Equivalently, the net work done by the object is positive.

Remember that the Wby object = -(Won on object)

What you are probably confusing is the concept that energy cannot be created or destroyed. Energy is merely transferred from one form to another (e.g. kinetic into potential, or kinetic into heat). Work is merely the term used to describe the magnitude of the energy transferred.


Edit: Allow me to elaborate further on what was incorrect in your post. The statement that "If you were to sum up all the heat that was lost as the ball slowed down, it would be equal to the kinetic energy the ball had inititally" is correct. The problem came with your next statement: "work wasn't being done on the ball (which would increase total energy) nor was work being done by the ball (which would decrease its total energy)". The frictional force does negative work on the ball, so it does decrease the ball's kinetic energy (and its total mechanical energy). An equal statement would be to say that the ball does positive work on its surroundings.

Da Tu Cha... aren't you tired of being right all the time

 

[Rofeh20]

you're quite right, friction is considered negative work. I think the concept of "negative work" is confusing in and of itself and it's better to consider the ball doing work on its surroundings than something doing negative work on the ball. To say "negative work is done on the ball" creates an exception that doesn't need to exist.

Quite simply if an object does work it loses energy and if an object has work done on it it gains energy (I know this isn't a point you're arguing but it is a point I'd like to make). Just as you pointed out, you could easily say an object has negative work done on it but I think it's conceptually wrong to think that this is what is happening. Obviously that's my opinion which counts for very little =).

Thanks for the clarification though. To take it a step further, any time a system loses energy to heat is the system doing work on its surroundings?

Yes, but we could discuss other models beyond the conversion of kinetic energy into heat. For example, a gas that undergoes adiabatic expansion can be said to be doing positive work on its surroundings, or equivalently it could be said that negative work is being done on the gas. I concur with your sentiment that the latter statement is confusing, but it is merely a game of semantics. As long as we understand that work is transfer of energy, whether it is positive or negative merely depends upon the perspective.

 

[Rofeh20]

Da Tu Cha... aren't you tired of being right all the time

Simma Down Na!!

And read The New Yorker. One other poster had the interesting idea to flip to the back of that magazine and read all of their music & theatre reviews. I liked that idea.

 

[BlackSails]

Actually, that's not correct. Via the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy.

W-KE theorem only applies in conservative systems, which do not include friction.

 

[Rofeh20]

W-KE theorem only applies in conservative systems, which do not include friction.

Hi BlackSails, that is not true. The work-Energy theorem applies when all of the energy transfer is related to a change in an object's kinetic energy. In the case we were discussing (and in the simplified physics models the AAMC will use), we are presuming that the only energy transfer that is occurring is kinetic energy into heat.

What you may be thinking of is the conservation of mechanical energy. Total mechanical energy is conserved when there are no dissipative forces present.

 

[BlackSails]

Hi BlackSails, that is not true. The work-Energy theorem applies when all of the energy transfer is related to a change in an object's kinetic energy. In the case we were discussing (and in the simplified physics models the AAMC will use), we are presuming that the only energy transfer that is occurring is kinetic energy into heat.

What you may be thinking of is the conservation of mechanical energy. Total mechanical energy is conserved when there are no dissipative forces present.

Simple counter example:

I strike a cymbal with a mallet. Part of the work I do goes into kinetic energy, part goes into sound.

Another counter example:

I mash a piece of silly putty with a hammer. None of the work I do goes into kinetic energy, it all goes into deformation (and some other minor modes.)


Just because the AAMC says it, doesnt make it true.

Edit: Yes, I know this is an MCAT forum. But that is no reason to spread falsehoods.

 

[Rofeh20]

Simple counter example:

I strike a cymbal with a mallet. Part of the work I do goes into kinetic energy, part goes into sound.

Another counter example:

I mash a piece of silly putty with a hammer. None of the work I do goes into kinetic energy, it all goes into deformation (and some other minor modes.)


Just because the AAMC says it, doesnt make it true.

Edit: Yes, I know this is an MCAT forum. But that is no reason to spread falsehoods.

Perhaps you didn't read my post carefully. You are not providing counter-examples, but more examples of when the work-energy theorem doesn't apply. There's no "flasehood" in the work-energy theorem, or in how I explained in it. I'm not sure why you have such a beef with the narrow set of circumstances under which it applies, but all I'm doing is helping folks understand how the AAMC tests physics on the MCAT. If you are uncomfortable with the AAMC's reductionist approach to physics I would suggest that you take that up with them.

 

[BloodySurgeon]

W-KE theorem only applies in conservative systems, which do not include friction.

I have only one thing to say....

Simma Down Na!!

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http://profile.imeem.com/bPIKf/video/qumxZ3LR/simmer_down_na/ <---CLICK URL

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[BloodySurgeon]

You can't find this video anywhere else just to let you know...

 

[BloodySurgeon]

my favorite is http://snltranscripts.jt.org/99/99lsimmer.phtml but I can't find it anywhere.