The work done by gravity on a ball thrown vertically upward

[Edel Crine]

Homework Statement

In a presence of air resistance, analyze the work done by gravity of all travel (ascend and descend) a ball thrown vertically upward.

Relevant Equations

w=f*d

It was just a conceptional question.
I thought that the positive work is done while it arises and negative work is done while it falls. For the effect of air resistance, the work done on each case would be less than the work done in absence of air resistance. Is there any other things that I can consider...? I just wanted to ask about it!
Thank you!

 

[HallsofIvy]

I'm not sure what you are trying to say. There is "positive" work done on the ball by whatever causes it to go upward to begin with. There is "negative" work done on the ball, as it is going first up and then down, by gravity and air resistance.

 

[etotheipi]

It's probably best that you look at all the forces individually. Start off with just the weight. The instantaneous power of a force, the instantaneous work done by that force per unit time, equals $\mathbf{F} \cdot \mathbf{v}$. If the force is in the same direction as the velocity, the power of that force is positive. If it is in the opposite direction, the power of that force is negative.

With that in mind, what is the sign of the work done by the weight force i) when the ball is rising and ii) when the ball is falling? Then, you can look at the other forces that act during different stages of the motion and try to apply signs the work they do.

Does this help to answer your question? And sorry if I misinterpreted it!

 

[Edel Crine]

I'm not sure what you are trying to say. There is "positive" work done on the ball by whatever causes it to go upward to begin with. There is "negative" work done on the ball, as it is going first up and then down, by gravity and air resistance.

Yes, I agree with you.
I think I wanted to say that
If we consider the work done by gravity only in the presence of air resistance, the total work done by gravity would be zero? What is the difference with the work done by gravity without air resistance??

 

[Edel Crine]

It's probably best that you look at all the forces individually. Start off with just the weight. The instantaneous power of a force, the instantaneous work done by that force per unit time, equals $\mathbf{F} \cdot \mathbf{v}$. If the force is in the same direction as the velocity, the power of that force is positive. If it is in the opposite direction, the power of that force is negative.

With that in mind, what is the sign of the work done by the weight force i) when the ball is rising and ii) when the ball is falling? Then, you can look at the other forces that act during different stages of the motion and try to apply signs the work they do.

Does this help to answer your question? And sorry if I misinterpreted it!

It did! 'Appreciate your help!

 

[etotheipi]

If we consider the work done by gravity only in the presence of air resistance, the total work done by gravity would be zero? What is the difference with the work done by gravity without air resistance??

You throw a ball up, and it travels a distance $h$ to the highest point in its motion. What is the work done by gravity? It falls down again a distance $h$, and you catch it at the same position that you threw it. What is the work done by gravity for this second stage of the motion? Then, what's the total work done by gravity for the whole trip? Does this number depend on air resistance?

You might find the notion of a conservative force interesting...

 

[Edel Crine]

You throw a ball up, and it travels a distance $h$ to the highest point in its motion. What is the work done by gravity? It falls down again a distance $h$, and you catch it at the same position that you threw it. What is the work done by gravity for this second stage of the motion? Then, what's the total work done by gravity for the whole trip?

What do you notice about this result? You might find the notion of a conservative force interesting...

They have same magnitude, but opposite direction. So the total work done by gravity will be zero (while the positive work done in the first stage and negative work done in the second stage), no matter of the air resistance?

 

[etotheipi]

They have same magnitude, but opposite direction. So the total work done by gravity will be zero (while the positive work done in the first stage and negative work done in the second stage), no matter of the air resistance?

That's the right answer, but there's one part of your explanation that's incorrect. When the ball is moving upward, what's happening to its speed (ignore air resistance for now)? So is it really positive work being done by gravity during this stage of the motion?

It might be helpful to draw a force diagram. Look at the direction of weight as compared to that of the velocity vector, for both stages of the motion. Remember what was said earlier about the sign of the dot product.

 

[Edel Crine]

That's the right answer, but you've gotten a little muddled. When the ball is moving upward, what's happening to its speed (ignore air resistance for now)? So is it really positive work being done by gravity during this stage of the motion?

It might be helpful to draw a force diagram. Look at the direction of weight as compared to that of the velocity vector, for both stages of the motion. Remember what was said earlier about the sign of the dot product.

OH my, I typed wrongly...hhh yes negative work during rising and positive work during falling

 

[Edel Crine]

That's the right answer, but there's one part of your explanation that's incorrect. When the ball is moving upward, what's happening to its speed (ignore air resistance for now)? So is it really positive work being done by gravity during this stage of the motion?

It might be helpful to draw a force diagram. Look at the direction of weight as compared to that of the velocity vector, for both stages of the motion. Remember what was said earlier about the sign of the dot product.

Also, will the less work be done in a case with air resistance than the case without it??
My apology for too many questions...

 

[etotheipi]

Also, will the less work be done in a case with air resistance than the case without it??
My apology for too many questions...

Make sure to be specific. The power of the force of gravity is $m\mathbf{g} \cdot \mathbf{v}$, and does this depend explicitly on the force of air resistance? Hopefully the answer is 'no'. What you've actually discovered with the zero work done is the motivation for introducing potential energies, since gravity is a conservative force which means the amount of work it does depends only on the start and end points of the path the object takes.

As for air resistance, in a fluid the drag is often modeled as a v-squared force, i.e. $\mathbf{F}_D = -k|\mathbf{v}|\mathbf{v}$. Hopefully you can see that it always has negative power, i.e. $P = \mathbf{F} \cdot \mathbf{v} = -k|\mathbf{v}| \mathbf{v} \cdot \mathbf{v} = -k|\mathbf{v}|^{3}$, which sort of makes sense since we expect it to be taking energy out of the system by acting to reduce the relative motion of the ball and the air!

It's important to specify which forces you're talking about. The work done by gravity doesn't depend on air resistance, since by definition it only considers the gravitational force. The work done by air resistance doesn't depend on the force of gravity, etc. You can treat all of these works separately.

But you might be interested in the total work done, since that's a useful quantity that you might equate to the change in kinetic energy. Luckily the total work done, or the work done by the 'resultant force', is the sum of the works done by all of the individual forces (can you prove this?). In that case, yes, the total work done by the resultant force is less positive in the presence of air resistance (can you see why?).

 

[Edel Crine]

Make sure to be specific. The power of the force of gravity is $m\mathbf{g} \cdot \mathbf{v}$, and does this depend explicitly on the force of air resistance? Hopefully the answer is 'no'. What you've actually discovered is the motivation for introducing potential energies, since gravity is a conservative force which means the amount of work it does depends only on the start and end points of the path the object takes.

As for air resistance, in a fluid the drag is often modeled as a v-squared force, i.e. $\vec{F}_{D} = -k|\mathbf{v}|\mathbf{v}$. Hopefully you can see that it always has negative power, i.e. $P = \mathbf{F} \cdot \mathbf{v} = -k|\mathbf{v}|\mathbf{v} \cdot \mathbf{v} = - k|\mathbf{v}|^{3}$, with sort of makes sense since we expect it to be taking energy out of the system by acting to reduce the relative motion of the ball and the air!

It's important to specify which forces you're talking about. The work done by gravity doesn't depend on air resistance, since by definition it only considers the gravitational force. The work done by air resistance doesn't depend on the force of gravity, etc. You can treat all of these works separately.

But you might be interested in the total work done, since that's a useful quantity that you might equate to the change in kinetic energy. Luckily the total work done, or the work done by the 'resultant force', is the sum of the works done by all of the individual forces (can you prove this?). In that case, yes, the total work done by the resultant force is less positive in the presence of air resistance (can you see why?).

That's perfect... I really appreciate! Have a great rest of the day!

 

[etotheipi]

That's perfect... I really appreciate! Have a great rest of the day!

Thanks , glad you liked it. If you come up with any more questions then do please ask!