How much work is done to hold a book at a constant height?

[ktsa]

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

 

[haruspex]

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

Per forum rules, you must show an attempt.
Explain why you think it could be zero.

 

[ktsa]

I think it could be zero because no displacement done
But I am not sure of that.
Can you help me?

 

[DaveE]

Can you tell us how work is defined in your course materials? Then evaluate each part of the appropriate equation. This should help you be sure. The problem isn't just to get the right answer, it's to understand why that answer is correct.

An example:
How much power (P, in Watts) is dissipated in a 20 Ohm resistor that has 10 Volts across it?
I know that the power is defined as P = V⋅I. I also know V = 10 Volts.
To find P, we also need to know the current through the resistor (I, in Amps). For that I know that Ohm's law defines the current as I = V/R. Since I know V =10 Volts and R = 20 Ohms, the current I = 10V/20Ω = 1/2 Amp.
So, back to P = V⋅I = 10 Volts ⋅ 1/2 Amp = 5 Watts.

And... yes, I am sure because I trust in each step of the process. I think you might not be sure because you are skipping steps in your analysis.

 

[ktsa]

In fact, in this problem, I know that work can be calculated using the formula of
w= F d cos theta
W= KEf- KEi
W= EPf- EPi........ and so one.
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.

Thank you

 

[DaveE]

why it is not possible to use the change in potential energy to fing the work done when holding the book at that height.

What is the change in potential energy? What is the change in kinetic energy?

 

[DaveE]

W= KEf- KEi
W= EPf- EPi........ and so one.

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.

Energy can appear in different forms. It is all of the energy (the sum) that is conserved (unchanged without work being done).

 

[ktsa]

Ok. Thank you

 

[Lnewqban]

...
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 believe that the key is in the question:
“how much work is done to hold a 0.80 kg book at a height...”

How did the book get there seems to be irrelevant in this case.
Potential energy is always a relative concept, and we always need to consider a change in height.

 

[kuruman]

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 ?

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.

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.

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.

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.

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, $W_{net}=W_{\!g}+W_T=mgh.$

 

[DaveE]

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.

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, $W_{net}=W_{\!g}+W_T=mgh.$

Yes. I shouldn't have said that. The work-energy theorem...