Work Energy Theorem

[CardiacArrest]

A crane lifts a 1000-kg steel beam off the ground, and sets it down on scaffolding 100 meters off the ground. All of the following are true EXCEPT:

A. The net work done on the steel beam is 9.8 x 10x5 j
B. The net work done on the steel beam is 0j
C. The magnitude of the work done on the steel beam by gravity is 9.8 x 10x5 J
D. The magnitude of the work done on the steel beam by the crane is 9.8 X 10x5J

The correct answer is A. Can anyone explain to me why 0J of work have been done? I know is work is equal to the change in kinetic energy, but does this mean that since the steel beam was at rest on the ground, and is eventually at rest off the ground, there is no work done?

Thanks!

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[lorenzomicron]

So, if they ask for network, we are asking for the work done in both phases of the object's motion. So, the first phase is lifting it, which uses some energy, given by choice D. Then, it is put down, which is work done, according to choice C. Adding these, gets you zero as well because the works are the same magnitude but in opposite directions.
However, more intuitively, the object gains energy, the loses the same energy, as its kinetic energy ends at zero. thus, there is no work.

 

[CardiacArrest]

Thanks!

Well taking the question a bit further, so would the work done be 9.8 x 10^5 J if the steel beam was just lifted off the ground and only hanging there? Because gravity would be pulling it down, right?

 

[phltz]

The idea here is that the crane is doing positive work on the beam. The crane is exerting an upwards force on the beam as the beam moves in the upwards direction. Work is F (dot) displacement, so this is positive work.

Meanwhile, gravity is exerting a downwards force on the beam as the beam moves in the upwards direction, which is negative work.

The total work done on the beam is 0. Thus, the beam has the same energy at the end as at the beginning. Note that in this case we're only considering the kinetic energy of the beam. The beam begins and ends at rest, so the kinetic energy doesn't change.

You're concerned about potential energy, and that's fair. What you need to understand is that from this perspective, the gravitational potential energy is not considered to belong to the beam.

Think of it this way: The beam has been raised through the gravitational field, and in doing so, gravity has done negative work. At any point, we could nudge the beam off the edge of the roof, at which point gravity would undo that negative work (which means gravity would do positive work), accelerating the beam downwards. So the potential energy is really just the potential to have work (energy) done by gravity. This isn't really energy that the beam has, it's energy that the beam could potentially get.

We sometimes go back and forth in which objects we consider potential energy to be "stored in", which makes this a more confusing question. For what it's worth, this question seems a bit more subtle and ambiguous than the questions they put on the official AAMC practice MCATs.

 

[fizzgig]

physics is kicking my ass all over town.

is it correct to think of this problem this way? the crane is doing work as it accelerates the beam upward (change in KE) and gravity does the same amount of work to decelerate it when the beam stops moving??

i know gravity is at work (for lack of a better phrase) the whole time, but unless it is overcoming the crane (or the crane is overcoming gravity), there's no net force so no movement, no KE change, and so no Work?

how does this fit in with W=Fd? i mean, the beam moved. is it Xwork when it's accelerating up then -Xwork when it's slowing down, and those F*d amounts cancel anyway?

 

[RogueUnicorn]

physics is kicking my ass all over town.

is it correct to think of this problem this way? the crane is doing work as it accelerates the beam upward (change in KE) and gravity does the same amount of work to decelerate it when the beam stops moving??

i know gravity is at work (for lack of a better phrase) the whole time, but unless it is overcoming the crane (or the crane is overcoming gravity), there's no net force so no movement, no KE change, and so no Work?

how does this fit in with W=Fd? i mean, the beam moved. is it Xwork when it's accelerating up then -Xwork when it's slowing down, and those F*d amounts cancel anyway?

that's the way i conceptualize it. crane works to move it up, gravity works to stop it. the crane work is W=Fd, while gravity's work is W=-Fd, since the movement is opposite to the direction of force applied. again, net is zero. there is a gain of potential energy, since gravity is a conserved force and it did negative work. any time gravity does negative work, you are gaining potential energy. the work from the crane is 'lost,' since it's not a conserved force.

i could be off here but that's how i view it.

 

[Jay2910]

Does this mean to say, that a guy that lifts a barbell from ground to a certain distance "d" and holds it there has also done no work?
If he is likened to the crane . .than he is providing kinetic energy to something that has 0 potential energy right? and if we were to let the barbell go then the gravitational potential energy would be doing positive work right?

Why is the crane here doing no work? is it because its setting something down?

 

[Patassa]

There's a lot of confusing things being said about this problem, hopefully this will be more clear and make sense.

This problem is solely about potential energy, not kinetic.
Don't forget the work energy theorem means work is also equal to the change in potential energy, not just the change in kinetic enery. W= mg(delta h)

Wcrane = 1000*9.8 * 100 Choice D

-Wgravity = 1000*9.8 * 100. Choice C (note the word magnitude, sign doesn't matter here)

Wnet = Wcrane + Wgravity = 0 Choice B (note the word net, sign makes all the difference here)

That's it, there is no other work being done and kinetic energy change is irrelevant because it goes back to zero, but that doesn't not mean work wasn't done or that the beam now has no energy.

The beam most certainly now has potential energy, mgh, because of the work done by the crane to get it there. Don't short change the crane, man.

Hope this helps.

 

[Patassa]

Does this mean to say, that a guy that lifts a barbell from ground to a certain distance "d" and holds it there has also done no work?

He did work to get it there, but he does no work to hold it there. Subtle but important difference.

Another important distinction is to keep the objects straight, the "who did what" here.

The man did work.
Gravity did work.
The weight had work done ON it.

 

[mehc012]

Does this mean to say, that a guy that lifts a barbell from ground to a certain distance "d" and holds it there has also done no work?
If he is likened to the crane . .than he is providing kinetic energy to something that has 0 potential energy right? and if we were to let the barbell go then the gravitational potential energy would be doing positive work right?

Why is the crane here doing no work? is it because its setting something down?

He would have done work...but gravity would have done an equal amount of work in the other direction (negative work) and so the net work done on the barbell would be zero.

Similarly, the crane is doing work...it's just being offset by the work gravity does.