Physics question about work/energy.

[willthatsall]

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I came across this question the other day and I can't figure out why the answer was what it was. Here's the deal: There is a heavy beam laying on the ground at a construction site. A crane lifts the beam off the ground and sets it down on a platform 100m off the ground. What is the work done on the beam?

I thought that the work done was equal to the potential energy of the beam at its new height, but the answer stated that the work done was 0, because the work done by gravity offset the work done by the crane. How can this be true, since the beam went from having no energy, to having potential energy? Doesn't there have to be some net work involved?

 

[MsEvolution]

Will, I would have answered as you did... if work equals the distance moved parallel to the force exerted... well... it should be equal to the 100m distance... so, if you're wrong, you're not alone. If someone figures this out, let me know because I want to be sure I have this concept down right!

 

[willthatsall]

Maybe the net work is related to the change in kinetic energy only? The kinetic energy doesn't change so I guess that would make the net work equal to 0.

Edited: I think that's it. Work-energy theorem states: W = delta K

 

[MsEvolution]

I've got my physics textbook with me, but the boss is here, so I can't whip it out right now! 😛

 

[midwest_turtle]

yeah, thats what my textbook says. just w=delta k

 

[willthatsall]

Interestingly, I actually got the problem right because it was one of those "all of the following are true, EXCEPT" questions, and the choices were like:

A: The net work is equal to 1 X 10^6 J
B: The net work is equal to 0 J
C: The net work done by the crane is equal to 1 X 10^6 J
D: The net work done by gravity is equal to 1 X 10^6 J

Obviously if 3 of those choices are true, then A has to be the false one. Just goes to show you can get some questions right even when you have no idea why.

 

[midwest_turtle]

Interestingly, I actually got the problem right because it was one of those "all of the following are true, EXCEPT" questions, and the choices were like:

A: The net work is equal to 1 X 10^6 J
B: The net work is equal to 0 J
C: The net work done by the crane is equal to 1 X 10^6 J
D: The net work done by gravity is equal to 1 X 10^6 J

Obviously if 3 of those choices are true, then A has to be the false one. Just goes to show you can get some questions right even when you have no idea why.

hopefully there will be alot of those come aug 14

 

[Assembler]

I don't get this. When going from 0m to 100m, the object loses kinetic energy while it gains potential energy. The PE should be +mgh, so the work should be W = -mgh. Also, the work done by the force is in the opposite of gravity, so theta = 180 degrees, thus W=-Fd = -mgh still. Doesn't make sense.

W = delta KE would work only if the speeds were the same (final and initial), or if they were both zero. I dont think that works.

 

[liverotcod]

It's a flawed question, IMHO. In the usual frame of reference used on the surface of the earth, work was done on the beam, because its potential energy was increased. The net energy of the system is unchanged, though.

 

[N-toxicologist]

I came across this question the other day and I can't figure out why the answer was what it was. Here's the deal: There is a heavy beam laying on the ground at a construction site. A crane lifts the beam off the ground and sets it down on a platform 100m off the ground. What is the work done on the beam?

I thought that the work done was equal to the potential energy of the beam at its new height, but the answer stated that the work done was 0, because the work done by gravity offset the work done by the crane. How can this be true, since the beam went from having no energy, to having potential energy? Doesn't there have to be some net work involved?

Does the beam really have PE, if was lifted onto a platform (which just happens to be 100m off the ground)? As in, if you let it go, it doesn't fall, right? PE=0 ? Somebody set me straight if my reasoning is off.

 

[willthatsall]

Does the beam really have PE, if was lifted onto a platform (which just happens to be 100m off the ground)? As in, if you let it go, it doesn't fall, right? PE=0 ? Somebody set me straight if my reasoning is off.

Like liverotcod was saying, in the normal frame of reference used on earth, yes it does have PE, mgh.

 

[N-toxicologist]

And how was it different in the question you cited? I understand how if B, C, and D were correct it would make A wrong (eg, the correct answer in this case), but I'm not getting the physics...

 

[Assembler]

Does the beam really have PE, if was lifted onto a platform (which just happens to be 100m off the ground)? As in, if you let it go, it doesn't fall, right? PE=0 ? Somebody set me straight if my reasoning is off.

Yes, but it was lifted off the ground first. The potential energy of the system is relative. At the ground, you could say the potential energy = 0. As you move to a greater height however, it will gain potential energy (energy by virtue of its position).

Gravitational PE only has to do with (vertical) height. If the object were to have moved horizontally, PE = 0 in all cases.

 

[Assembler]

Now that I look more at this problem...I think it is possible that the work done by the crane offset that of the gravity. Didn't the crane have to attach to a beam using some kind of support that would act for tensile strength?

 

[shad420w]

The question is right.... Because it never asks for the work done by the crane nor gravity.... so it appears to be asking for total work, and the work done by the crane is equal but opposite of the work of gravity.
Wcrane = -Wgravity

Thus Wnet = Wcrane + Wgravity = 0

This maybe what they are getting at??

 

[MoosePilot]

I would have calculated it as w=mgh. I don't buy the explanation that the two works cancel, something has definitely happened. You could tie the beam to a rope, drop it off the platform, and harness it's PE to do work, so where did that energy come from? It came from the work done on the beam.

 

[aditi]

Work = FdCos(theta) where d is the displacement and theta is the angle between F and d. In this case, theta is 0, so Work = Fd. F should be equal to the weight of the beam. So, with respect to the beam, there is net work done.

 

[MoosePilot]

Work = FdCos(theta) where d is the displacement and theta is the angle between F and d. In this case, theta is 0, so Work = Fd. F should be equal to the weight of the beam. So, with respect to the beam, there is net work done.

F=mg
d=h
Fd=mgh

This is the same result a lot of us get, but another way to get there. I totally agree.

 

[GBFKicks]

Well, as other people have mentioned, I believe the answer lies in the way the question is worded. It doesn't ask for the work done by the crane or the work done by gravity--rather, it asks for the work done on the beam (net work of the system). I think one way of looking at it, is to say that the F in Fdcos(theta) is the net force of the system, which is zero. Therefore the W in the system is 0.
You guys were right and if the question had asked for the work done by the crane or the work done by gravity, F would change.

Correctly me if I'm wrong here... I hate physics in every way possible

 

[MoosePilot]

Well, as other people have mentioned, I believe the answer lies in the way the question is worded. It doesn't ask for the work done by the crane or the work done by gravity--rather, it asks for the work done on the beam (net work of the system). I think one way of looking at it, is to say that the F in Fdcos(theta) is the net force of the system, which is zero. Therefore the W in the system is 0.
You guys were right and if the question had asked for the work done by the crane or the work done by gravity, F would change.

Correctly me if I'm wrong here... I hate physics in every way possible

I don't think so, because the beam moved. If you define the system as just the beam, it picked up some potential energy from somewhere, which is the work done on it. Just my opinion, I'd be willing to be schooled 🙂

 

[RayhanS1282]

That's a pretty f*$ked up question right there. I agree with the majority of the posters and say that the beam has potential energy because its position changed. So, there was work being done on the beam by the crane. The only thing that, to me at least seems 0 is the forces acting on the beam...hmmm maybe they were asking the question from that perspective? If net force on the beam is zero therefore W=FD, so, W=0. Is this even remotely correct? I can't solve energy problems and the MCAT is around the corner....good times!!!

 

[captain hazel]

it's weird because it's forcing you to set up your system in a specific way; usually you have a choice.

if you consider the system to be the beam & the earth (so the crane is the external force acting on the system):

the PE of the system increases as the crane lift the beam away from the earth. deltaPE=mgh, which is the amount of work the crane did on the system (the beam & the earth).


but...this question is asking for the work done on the beam only. so you it's best to consider the system to be the beam alone (so the earth & the crane are the external forces acting on the system):

like you said, W=deltaKE, and deltaKE=0. you did positive work on the beam, gravitity/earth did negative work on the beam, so the net work is 0.

PE isn't an especially meaningful value in this system because your system is just one object--the beam. external Potential Energies are between two objects (like the earth & an object, or between two planets, or between two charges, etc.)



the best way to do this (or any) problem:
the fastest way that gets you to the right answer. i.e. exactly the way you did it, will.

😀

 

[MoosePilot]

I still don't agree, because the beam is in Earth's gravity field. There are so many ways to hook the beam up to do work as it falls that it couldn't have done before. That ability had to come from somewhere.

Maybe I'm just unclear on this type of problem, but I don't think so.

 

[pathdr2b]

Well, as other people have mentioned, I believe the answer lies in the way the question is worded. It doesn't ask for the work done by the crane or the work done by gravity--rather, it asks for the work done on the beam (net work of the system).

👍
Agreed! The key word is work ON the beam, not work BY the beam. Thank you, Thank you for EK-Physics as problems like this are the norm!

 

[aditi]

Interestingly, I actually got the problem right because it was one of those "all of the following are true, EXCEPT" questions, and the choices were like:

A: The net work is equal to 1 X 10^6 J
B: The net work is equal to 0 J
C: The net work done by the crane is equal to 1 X 10^6 J
D: The net work done by gravity is equal to 1 X 10^6 J

Obviously if 3 of those choices are true, then A has to be the false one. Just goes to show you can get some questions right even when you have no idea why.

In this question, was the beam dropped by the crane after it was lifted up to a certain height? If that was the case then choice A would be the correct answer; if not choice a and c should true and b and d should be false.

Does this make sense or am I missing something?

Thanks

 

[MoosePilot]

Interestingly, I actually got the problem right because it was one of those "all of the following are true, EXCEPT" questions, and the choices were like:

A: The net work is equal to 1 X 10^6 J
B: The net work is equal to 0 J
C: The net work done by the crane is equal to 1 X 10^6 J
D: The net work done by gravity is equal to 1 X 10^6 J

Obviously if 3 of those choices are true, then A has to be the false one. Just goes to show you can get some questions right even when you have no idea why.

Ok, I give in. I think B is right, too, now.

A related question, though. If you look at the beam halfway up to the platform, while it's still moving, what is the work done by the crane, what is the work done by gravity, and what is the net work? It's moving, so if you use Delta KE you'll get a positive answer, but I don't see how it's significantly different from the original question.