For this question, I understand the work done by force F is 200 (mgh), but I assumed that since there are two pulley systems, the work should be divided by 2 and thus be 100, but it is not the correct answer. Could someone explain?
Physics Discrete
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For this question, I understand the work done by force F is 200 (mgh), but I assumed that since there are two pulley systems, the work should be divided by 2 and thus be 100, but it is not the correct answer. Could someone explain?
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Is the answer not just 200 J?
The work is the same, simply the change in potential energy of the mass = mgh = 200 J.
The difference with the pulleys is that the force, F, is less than if you lifted it straight up.
If lifted over a single pulley, then a force of Mg would be required. If lifted with this 2 pulley system, a force of 0.5Mg is required.
The key point is that the work performed is the same.
Also, this is the wrong forum for this question.
The work is the same, simply the change in potential energy of the mass = mgh = 200 J.
The difference with the pulleys is that the force, F, is less than if you lifted it straight up.
If lifted over a single pulley, then a force of Mg would be required. If lifted with this 2 pulley system, a force of 0.5Mg is required.
The key point is that the work performed is the same.
Also, this is the wrong forum for this question.
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You said you knew the work was 200J, then you divided it by the number of pulleys.
You can divide the force required by the number of pulleys to express the mechanical advantage.
Work is work is work - it is independent of path and simply reflects the change in energy of the system.
The force depends on how you do it, in this case you use less force, but over a longer path.
You can divide the force required by the number of pulleys to express the mechanical advantage.
Work is work is work - it is independent of path and simply reflects the change in energy of the system.
The force depends on how you do it, in this case you use less force, but over a longer path.
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Cawolf:
I was thinking when I did it that the work would be Mgd(cosQ). Why wasn't the cos 60 taken into consideration? I realized that I did not fully get the concept there...
I was thinking when I did it that the work would be Mgd(cosQ). Why wasn't the cos 60 taken into consideration? I realized that I did not fully get the concept there...
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Work is simply the change in energy and is independent of path.
No matter how we lift or lower that mass, the work is equivalent to the change in potential energy.
This is known as the Work-Energy Theorem and therefore makes the angle of applied force irrelevant.
No matter how we lift or lower that mass, the work is equivalent to the change in potential energy.
This is known as the Work-Energy Theorem and therefore makes the angle of applied force irrelevant.
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Thanks a lot.
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@Cawolf is the man! He's spot on here.
Many times people limit their perspective of work on it just being equal to F*d*cos(theta), which of course is not wrong, but it's so, so necessary and beneficial to think of it in terms of changing energy (potential in this case).
The force equation for work functions fine in this situation. The angle does not matter since the cord that is actually causing the block to move is parallel to the the block's movement. Cosine of 0 degrees is 1.
Also using the potential energy equation mgh works just as well here, since again, work is the change in energy.
Many times people limit their perspective of work on it just being equal to F*d*cos(theta), which of course is not wrong, but it's so, so necessary and beneficial to think of it in terms of changing energy (potential in this case).
The force equation for work functions fine in this situation. The angle does not matter since the cord that is actually causing the block to move is parallel to the the block's movement. Cosine of 0 degrees is 1.
Also using the potential energy equation mgh works just as well here, since again, work is the change in energy.
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NextStepTutor and Cawolf:
Sorry to be a pest, but I still have a few questions...
Are you guy saying that the equation F*d*cos (theta), is valid? Is it just a matter of the angle theta? In this case cos 0 being 1. In other words, when would one know when to use work = F*d*cos(theta), as opposed to work = F*d? Is it just that work = F*d*cos(theta), in all case and we just need to be careful of the angle (theta)?
(Are you guy just totally confused by me trying to ask my question...LOL...
Sorry to be a pest, but I still have a few questions...
Are you guy saying that the equation F*d*cos (theta), is valid? Is it just a matter of the angle theta? In this case cos 0 being 1. In other words, when would one know when to use work = F*d*cos(theta), as opposed to work = F*d? Is it just that work = F*d*cos(theta), in all case and we just need to be careful of the angle (theta)?
(Are you guy just totally confused by me trying to ask my question...LOL...
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@MrMention
In the example posted, the pulleys (assumed ideal) simply change the direction of the applied force without adding mass or friction to the system. The angle given in that problem simply seeks to confuse people.
So Fdcos(θ) is valid but the distance will be 10m (twice the height the mass is raised), the Force will be 20 J (half of the force without the mechanical advantage of 2), and θ = 0 as the force is in the direction of displacement.
My suggestion is that for any work problem, the PE and KE calculations will be easier as there is no dot product involved. Since work is a scalar function, the absolute value will always be correct.
In the example posted, the pulleys (assumed ideal) simply change the direction of the applied force without adding mass or friction to the system. The angle given in that problem simply seeks to confuse people.
So Fdcos(θ) is valid but the distance will be 10m (twice the height the mass is raised), the Force will be 20 J (half of the force without the mechanical advantage of 2), and θ = 0 as the force is in the direction of displacement.
My suggestion is that for any work problem, the PE and KE calculations will be easier as there is no dot product involved. Since work is a scalar function, the absolute value will always be correct.
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Thanks, I appreciate it.
