The discussion revolves around calculating the work done to hold a 0.80 kg book at a height of 1.3 m for 5.0 seconds. Participants are exploring the concept of work in the context of physics, particularly focusing on whether the work done is zero due to lack of displacement.
There is an ongoing exploration of different interpretations regarding the work done by various forces, including gravity and the force exerted by the hand holding the book. Some participants have provided guidance on evaluating the definitions of work and potential energy, while others express uncertainty about the implications of these definitions.
Participants note that the problem may be ambiguous as it does not specify which force is doing the work, leading to different interpretations of the scenario. The discussion also highlights the importance of understanding the relationship between work and energy in conservative systems.
Per forum rules, you must show an attempt.ktsa said:Homework Statement: how much work is done to hold a 0.80 kg book at a height of 1.3 m above ground for 5.0 s ?
Relevant Equations: Please can anyone help me with problem? I don't know if the work is 0 J
Please can anyone help me with problem? I don't know if the work is 0 J
What is the change in potential energy? What is the change in kinetic energy?ktsa said:why it is not possible to use the change in potential energy to fing the work done when holding the book at that height.
Not quite correct. This really should be one equation W = (KEf + EPf) - (KEi + EPi). There are circumstances, like pendulums and orbits where the change in kinetic energy is exactly compensated by an opposite change in potential energy, with no net work done. Those formulas only work if you specify that the other sort of energy doesn't also change.ktsa said:W= KEf- KEi
W= EPf- EPi........ and so one.
I believe that the key is in the question:ktsa said:...
But I ask for this because one other tutor said that the answer is 0 J because there is no displacement done, which can also be true. That is why I was wondering why it is not possible to use the change in potential energy to fing the work done when holding the book at that height.
I think this question is just a tad ambiguous because it does not specify the force that does the work. It could be the work done by gravity or by the hand holding the book or by the net force. Of course, in this case it doesn't matter because, as has already been argued, the displacement is zero hence neither force does work on the book and adding the two to get the work done by the net force yields zero.ktsa said:Homework Statement: how much work is done to hold a 0.80 kg book at a height of 1.3 m above ground for 5.0 s ?
Here is where it becomes important to know which force does the work. The change in potential energy is by definition the negative of the work done by a conservative force. You can use the definition here if the problem specifically asked you to find the work done by gravity but not if it asked you to find the work done by the hand.ktsa said:That is why I was wondering why it is not possible to use the change in potential energy to fing the work done when holding the book at that height.
I find this statement misleading. Take a simple pendulum that swings from maximum height ##h## to the lowest point of the motion. There are two forces acting on the bob:DaveE said:There are circumstances, like pendulums and orbits where the change in kinetic energy is exactly compensated by an opposite change in potential energy, with no net work done.
DaveE said:There are circumstances, like pendulums and orbits where the change in kinetic energy is exactly compensated by an opposite change in potential energy, with no net work done. Those formulas only work if you specify that the other sort of energy doesn't also change.
Yes. I shouldn't have said that. The work-energy theorem...kuruman said:I find this statement misleading. Take a simple pendulum that swings from maximum height ##h## to the lowest point of the motion. There are two forces acting on the bob:
1. Gravity that does work ##W_{\!g}=mgh##.
2. Tension that does zero work, ##W_T=0##, because the force is always perpendicular to the displacement.
We have a system that conserves mechanical energy yet the net work, i.e. the sum of the works done by all the forces, is non-zero, ##W_{net}=W_{\!g}+W_T=mgh.##