Net work?

[AsherLev]

I was doing Kaplan full length 7, and came across a question about net work..

the gist of the question was that a crane lifted a metal beam above the ground, and set it down 100m above original point. The answer asserted that Zero net work was done...asserting that while the crane did positive work to bring the beam up...that the beam did negative work on gravity...so in this case...wouldnt most all net work be zero...even if i lift a box up (positive work) negative work is done on gravity...canelling out to zero...so im confused...ie...will Net work EVER be done?

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[littlealex]

I think you need to type out the ACTUAL question, word by word, and the actual answer given, also word by word.

The reality of the situation is that net work in the universe is 0, but the net work done by the crane on the metal beam is positive, and the net work done by gravity on the beam is negative.

Net work can only be found when it is directly pertinent to a body applying the force.

 

[saidel273]

To my understanding, the net work done by conservative forces is zero. In your case, gravity is a conservative force so the net work will be zero.

However, if your force is non-conservative, then there will be a net work done on the system.

 

[reine1jb]

I was doing Kaplan full length 7, and came across a question about net work..

the gist of the question was that a crane lifted a metal beam above the ground, and set it down 100m above original point. The answer asserted that Zero net work was done...asserting that while the crane did positive work to bring the beam up...that the beam did negative work on gravity...so in this case...wouldnt most all net work be zero...even if i lift a box up (positive work) negative work is done on gravity...canelling out to zero...so im confused...ie...will Net work EVER be done?

you have to remember that force is a vector quantity. W=Fdcos theta
So you have to think of everything in terms of direction. As the problem states, the crane lifting the metal beam up is considered positive work because we assoicate the up direction as positive. Also, as the answer explains, the Force of gravity is pointing down, therefore, the work would be negative.

I think if I have this right, work is independent of the path taken and is only dependent on the difference between the start and the end point (think equipotential lines and potential difference bewteen each line) so in that light, if you end where you started you essential did no work as the answer to the problem states

 

[AsherLev]

50. 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 × 105 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 × 105 J.
D. The magnitude of the work done on the steel
beam by the crane is 9.8 × 105 J.

Answer Explanation:

50. A
All of the answer choices have to do with the work done on the steel beam as it is lifted 100
m off the ground. Let’s evaluate each answer choice:
A: What is the net work done on the steel beam? Since the change in kinetic energy of the
beam is zero, the work-energy theorem tells us that the net work done on the beam must be zero
as well. So this statement is false, making A the correct answer. Let’s check the other answers.
B: The net work done on the beam is zero, as we discovered above.
C: The work done on the beam by gravity is negative (the beam does work on the Earth),
because the direction of the force (downwards) is opposite the direction of displacement
(upwards). The magnitude of the work is mgh = (1000 kg)(9.8 m/s2)(100 m) = 9.8 × 105 J.
D: The work done on the beam by the crane is positive (force and displacement both point
up) and is also mgh = 9.8 × 05 J. Notice that the work done by gravity cancels the work done by
the crane; confirmation that the net work done on the beam is 0 J.

 

[saidel273]

I think if I have this right, work is independent of the path taken and is only dependent on the difference between the start and the end point (think equipotential lines and potential difference bewteen each line) so in that light, if you end where you started you essential did no work as the answer to the problem states

In general, I believe work is a path-dependent function. W = Fd and d is distance not displacement. So your total work will depend on the path.

I believe the exceptions are with conservative forces like gravity and electrostatic forces.