Work energy

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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 × 10^5 J.

B. The net work done on the steel beam is 0 J.

C. The magnitude of the work done on the steel beam by gravity is 9.8 × 10^5 J.

D. The magnitude of the work done on the steel beam by the crane is 9.8 × 10^5 J.

Answer: A.




Well this is disappointing. I thought I knew my work and energy well. Please, no test taking strat suggestions. I know I can eliminate C and D. I'm looking for a conceptual understanding.

Why is net work done on the steel beam 0 J? In lifting it, its potential energy increases. Is this not work done?

thanks.
 
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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 × 10^5 J.

B. The net work done on the steel beam is 0 J.

C. The magnitude of the work done on the steel beam by gravity is 9.8 × 10^5 J.

D. The magnitude of the work done on the steel beam by the crane is 9.8 × 10^5 J.
Click to expand...

Yeah that doesn't make any sense.
Energy was clearly imparted into the steel beam as it now has greater PE. And you can't do that without W.
 
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The only thing I can think of is that gravity is pushing it down (9.8*10^5) and the crane is lifting it up (9.8*10^5), which nets to 0. But that's net force, not net work, right?
 
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Net work is the work done by all the forces on the object. Net work is the same as the change in kinetic energy.

In this case W of work was done by the lift and -W by gravity, for a total of 0. Alternatively, since the speed is 0 before and after the move, the change in KE is zero and the net work is 0.


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MedPR said:
The only thing I can think of is that gravity is pushing it down (9.8*10^5) and the crane is lifting it up (9.8*10^5), which nets to 0. But that's net force, not net work, right?
Click to expand...

But you almost got there! 🙂

Net work will be net force times the displacement...


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I know you didn't want test taking strats but you can also eliminate B because if C and D are true, B has to be true also, because you can't add or subtract 9.8x10^5 from itself and get 9.8x10^5.
 
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milski said:
Net work is the work done by all the forces on the object. Net work is the same as the change in kinetic energy.

In this case W of work was done by the lift and -W by gravity, for a total of 0. Alternatively, since the speed is 0 before and after the move, the change in KE is zero and the net work is 0.


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I am here: http://tapatalk.com/map.php?q0zzt1
Click to expand...

Why is net work the same as change in kinetic energy?


kasho11 said:
I know you didn't want test taking strats but you can also eliminate B because if C and D are true, B has to be true also, because you can't add or subtract 9.8x10^5 from itself and get 9.8x10^5.
Click to expand...

that only makes sense if you understand the answer.
 
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Net work and work are hard to understand. Displace an object, and you will have done work on it. Displace an object, and then return it, and you would still have done work on it. In both case, if there is no change in the v of the object (or KE), net work is 0.

w = fd = mad = md dv/dt

if there was a change in v, net work would depend on that and d.
 
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So if i slide a box across a table, net work =0

Because frictional force and the force of me pushing the box is equal.
If there is KE then there is work netted.
If there is no KE then there is no work done.
 
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johnwandering said:
So just write down
Net force =KE?
Click to expand...

net force = change in KE. just just KE.

hellocubed said:
So if i slide a box across a table, net work =0

Because frictional force and the force of me pushing the box is equal.
If there is KE then there is work netted.
If there is no KE then there is no work done.
Click to expand...

i think so. it is similar to the original question because there, gravity did negative work and the movement upwards did positive work for a net = 0.

with pushing a box against a friction-full surface, the force we exert is some value but there is negative work done by the friction since the force is in the opposite direction and W = Fdcos(theta). That cos(theta), I think, is what I had neglected to consider the first time.

would like verification that i'm correct, though.
 
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Wait this still doesnt make sense.
Im sorry, i dont like to just write down facts if i dont understand them.

We are saying that y directional net force is zero, but how?
We have gravitaional force going down.
An we have lift force going up.

If the object moves up, lift force is clearly LARGER than gravitational force.
And W=Fx
Because lift and gravity force act across SAME distance, the work imparted via lift force should be greater than the work imparted via gravity.

The net work shouldnt be zero....

It would be zero if the box came down to the same genesis, but it does not. The net W by lift force is greater than the net W by gravity, because the box has been lifted a greater distance than fallen.
 
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Also, it seems this would violate the laws of thermodynamics

We created PE without any Net E input...
 
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net work is change in KE, not net force.

Yes, it's more complicated which is why in an intro class you'll only see it as work-kinetic-energy theorem without a proof. It's relatively straightforward to prove for a constant net force but it is still correct for any net force and path.

There are no problems with conservation of energy - the potential energy increase of the steel beam is the same as the work done by the crane - it is lost in someone way by the crane, most likely as electric or potential chemical energy.

If you want to go in the details, if you assume that the speed up is constant for most of the movement, the lift and weight will be the same. There will be a short period in the beginning when the lift will be larger than the weight which will accelerate the beam from 0 to some speed. There will be a similar situation at the top when the beam is decelerated to 0. There are not necessary equally long but it can be proven that the work done by each force cancels.
 
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milski said:
net work is change in KE, not net force.

Yes, it's more complicated which is why in an intro class you'll only see it as work-kinetic-energy theorem without a proof. It's relatively straightforward to prove for a constant net force but it is still correct for any net force and path.

There are no problems with conservation of energy - the potential energy increase of the steel beam is the same as the work done by the crane - it is lost in someone way by the crane, most likely as electric or potential chemical energy.

If you want to go in the details, if you assume that the speed up is constant for most of the movement, the lift and weight will be the same. There will be a short period in the beginning when the lift will be larger than the weight which will accelerate the beam from 0 to some speed. There will be a similar situation at the top when the beam is decelerated to 0. There are not necessary equally long but it can be proven that the work done by each force cancels.
Click to expand...

How is this problem any different from a problem that asks how much work is done by pushing a rock up an incline plane? You could do it by solving for the hypotenuse of the triangle (the length of the plane) and do W=Fd or you could just do mgh. It certainly doesn't come out to 0 though...
 
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MedPR said:
How is this problem any different from a problem that asks how much work is done by pushing a rock up an incline plane? You could do it by solving for the hypotenuse of the triangle (the length of the plane) and do W=Fd or you could just do mgh. It certainly doesn't come out to 0 though...
Click to expand...

Proper terminology helps a lot. When you talk about work, you should talk either about work done by force F on object X or about net work done on object X.

Pushing a rock up on an incline plane from rest at the bottom to rest on the top:
- net work done on the rock - 0
- work done on the rock by friction - x
- work done on the rock by the pushing force - y
- work done on the rock by gravity - z
 
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I think the trouble people are having is that there are two examples:

1.) a box is puhed horizontally from point 1 to point 2. No net energy is gained, no net W done.

2.) a box is lifted from point 1 to point 2. Mgh PE is gained.
Why are we considering an imperfect system where E is lost through mechanical and chemical E?
If it is not, then E is clearly gained.

I dont even see how its disputable that E is not gained by the use of a perfect crane. Because its quite clear.
Box initially E=0
Box lifted E=mgh
 
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milski said:
Proper terminology helps a lot. When you talk about work, you should talk either about work done by force F on object X or about net work done on object X.

Pushing a rock up on an incline plane from rest at the bottom to rest on the top:
- net work done on the rock - 0
- work done on the rock by friction - x
- work done on the rock by the pushing force - y
- work done on the rock by gravity - z
Click to expand...

Oh that's right. I remember being so confused about this until I did that one NOVA Physics problem with the old woman pushing her grocery cart up a hill. Nice work!
 
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johnwandering said:
I think the trouble people are having is that there are two examples:

1.) a box is puhed horizontally from point 1 to point 2. No net energy is gained, no net W done.

2.) a box is lifted from point 1 to point 2. Mgh PE is gained.
Why are we considering an imperfect system where E is lost through mechanical and chemical E?
If it is not, then E is clearly gained.

I dont even see how its disputable that E is not gained by the use of a perfect crane.
Click to expand...

There is no dispute about energy gained or lost. Just because energy was gained, does not mean that the net work is not zero.

The U=W+Q from thermo applies only the whole system - if you want it to hold, you need to include everything in the system, including the crane.
 
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If you want to go in the details, if you assume that the speed up is constant for most of the movement, the lift and weight will be the same. There will be a short period in the beginning when the lift will be larger than the weight which will accelerate the beam from 0 to some speed. There will be a similar situation at the top when the beam is decelerated to 0. There are not necessary equally long but it can be proven that the work done by each force cancels.
Click to expand...

this explains it quite well. thanks
 
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