Physics - From Acceleration, to Force, to Momentum, to Power, Oh My! (EK)

[SKaminski]

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Okay, So I think I may be having some issue's understanding some basic principals of physics, and I THINK it's rooted in some basic confusion about force/tension, as well as some confusion regarding conservative/non conservative forces, when which is being applied, and when to use which formula's.

All questions are either derived from my own imagination, or from Exam Krackers.

Question 1
My first question is a general one, and is from my own imagination.

Let's say I go sky diving, and I hit terminal velocity. Now, I know that gravity exerts a constant acceleration on me at 9.8 m/s^2. I also know that wind resistance is giving an acceleration in the OPPOSITE direction at 9.8m/s^2. This means that I am in equilibrium. Equilibrium equals no change in velocity, and no change in velocity equals no acceleration.

But... do I have an acceleration? Because I do. I have an acceleration of 9.8 m/s^2 downward, and I have an acceleration of 9.8 m/s^2 upward. Is this correct, or in correct?

Question 2
I think this one ties into my previous question regarding net acceleration, but is now stated in terms of Force.
EK 1001 Physics Problem, Question 264:
A 25 kg mass is lowered by a rope. If the velocity of the mass is decreasing at a rate of 5 m/s^2, what is the tension in the rope?

The credited answer is 180, because T = mg + ma. This makes sense if you draw the picture. You have mg (25 * 9.8) pulling down, and ma (25 * 5)pulling up, slowing down the decent. Easy peasy!

EK 1001 Physics Problem, Question 266:
A 32 kg mass is raised by a rope. If the velocity of the mass remains constant at 2 m/s, what is the tension in the rope?

The credited answer is 320N. Because T = mg.

This doesn't make sense to me in light of the question 264, and here's why.

In the previous problem, I used the opposing force + gravitational force to determine the tension. T = (25 * 9.8) + (25 * 5). In this problem, I'm not using the opposing force anymore. If I attempted to solve this problem with the same approach as I tried to solve the previous problem, it would be T = (25 * 9.8) + (25 * 9.8)

This is clearly not correct. The only reasonable explanation I can think of would be that we don't use the 'opposing force' at all. This doesn't seem correct though, because it states (in the answer book for question 264) that the answer is T = mg + ma. I was using 5 m/s^2 for my acceleration upward, and 9.8 m/s^2 for my acceleration downward. Total tension came out correctly. Should I have used 15 m/s^2 for my upward acceleration? That would give me an incorrect answer...

So, if a gravitational opposing force is factored into one answer when it is SLOWING DOWN the effect of gravity, why is a gravitational opposing force not factored in when it is NEGATING gravity? I'm so lost.

Question 3
Physics EK 1001, Question 263
A 10 kg mass hangs from a rope. A force is applied to the rope so that the mass is accelerated upward at 2 m/s^2. What is the tension in the rope.

Credited Answer: 120N.

If i were to nail a string to an overhang, and dangle a weight from it, it would have Tension t1. If I were to pick up that weight, and hold it next to the nail, the tension would be zero, because there would be no tension in the string (the string is slack!). If I were to, next, push the string upward so that it had an acceleration of 2 m/s^2 (and the mass weighed 10kg), shouldn't my Force be 2 m/s^2? Am I thinking about this mystical 'force' the wrong way?

This may contribute to my confusion in Question 2.

Removed Several Questions to Give them there own thread, as suggested by LoLCareerGoals

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And that's it! I know the post is lengthy, and I know it hits you like a wall of text, but dammit I'm feeling extremely perplexed, and would appreciate ANY direction ANY of you might be able to provide. Currently things seem so arbitrary for what forces you chose to include, and whether or not you calculate a problem with energy or work or power, and even on where you put the weights on the board! Agh!

If it helps at all (not trying to brag) I got 4.0's in both of my physics class in college (graduated 2 years ago). I don't know if i learned it a different way, or if I forgot everything, but this is really tripping me up.

Again, any insight ANY of you can offer is SO appreciated. Thanks!

 

[LoLCareerGoals]

Question 1
My first question is a general one, and is from my own imagination.

Let's say I go sky diving, and I hit terminal velocity. Now, I know that gravity exerts a constant acceleration on me at 9.8 m/s^2. I also know that wind resistance is giving an acceleration in the OPPOSITE direction at 9.8m/s^2. This means that I am in equilibrium. Equilibrium equals no change in velocity, and no change in velocity equals no acceleration.

But... do I have an acceleration? Because I do. I have an acceleration of 9.8 m/s^2 downward, and I have an acceleration of 9.8 m/s^2 upward. Is this correct, or in correct?

1 question per thread is the custom around here. Anyway on to q1:
You always have an acceleration, it may equal zero, but you sure do have it. Let's use more precise scientific language.
In your specific example, upon reaching terminal velocity using free fall terminal velocity definition where it is the velocity you reach when the drag (air resistance) equals gravitational force you have zero acceleration. Had you not had zero acceleration, there'd be nothing "terminal" about your velocity as it would still be changing.

Acceleration of an object is a net vector sum of multiple accelerations. One is up one is down, net is zero.

 

[LoLCareerGoals]

Question 2
I think this one ties into my previous question regarding net acceleration, but is now stated in terms of Force.
EK 1001 Physics Problem, Question 264:
A 25 kg mass is lowered by a rope. If the velocity of the mass is decreasing at a rate of 5 m/s^2, what is the tension in the rope?

The credited answer is 180, because T = mg + ma.

This is only true when the vector a is in the same direction as vector g.
I think a lot of your confusion stems from lack of understanding of vectors. Stop memorizing formulas like one above and understand Newton's first law:
vector sum of forces acting on an object = mass of an object x acceleration vector

This makes sense if you draw the picture. You have mg (25 * 9.8) pulling down, and ma (25 * 5)pulling up, slowing down the decent. Easy peasy!

EK 1001 Physics Problem, Question 266:
A 32 kg mass is raised by a rope. If the velocity of the mass remains constant at 2 m/s, what is the tension in the rope?

The credited answer is 320N. Because T = mg.

This doesn't make sense to me in light of the question 264, and here's why.

In the previous problem, I used the opposing force + gravitational force to determine the tension. T = (25 * 9.8) + (25 * 5). In this problem, I'm not using the opposing force anymore. If I attempted to solve this problem with the same approach as I tried to solve the previous problem, it would be T = (25 * 9.8) + (25 * 9.8)

Once again you are misapplying a formula which is nothing more than a very special case of Newton's first law. a of an object is zero here because its speed is not changing!
There are 2 forces acting on object: force of gravity (down) and force of tension (up).
Then vector sum of forces is:
mg - T = mx0 = 0, hence T = mg (if you want positive to be down) OR
-mg + T = mx0 = 0, hence T = mg (if want positive to be down) -> same answer.

This is clearly not correct. The only reasonable explanation I can think of would be that we don't use the 'opposing force' at all. This doesn't seem correct though, because it states (in the answer book for question 264) that the answer is T = mg + ma. I was using 5 m/s^2 for my acceleration upward, and 9.8 m/s^2 for my acceleration downward. Total tension came out correctly. Should I have used 15 m/s^2 for my upward acceleration? That would give me an incorrect answer...

So, if a gravitational opposing force is factored into one answer when it is SLOWING DOWN the effect of gravity, why is a gravitational opposing force not factored in when it is NEGATING gravity? I'm so lost.

See above.


For question 3, set up a vector equation yourself first. Label all quantities: T, m, g, a if applicable. What is a equal to from problem definition? Which way is your positive direction (imaginary reference axis)?

 

[SKaminski]

This is only true when the vector a is in the same direction as vector g.
I think a lot of your confusion stems from lack of understanding of vectors. Stop memorizing formulas like one above and understand Newton's first law:
vector sum of forces acting on an object = mass of an object x acceleration vector

Once again you are misapplying a formula which is nothing more than a very special case of Newton's first law. a of an object is zero here because its speed is not changing!

First, let me say that I appreciate you taking the time to answer my questions. Secondly, I wasn't fully aware of of the custom to post 1 question at a time. I tried that last time I had free time to post my questions, and I ended up creating 5 threads. 😛. I'll revert back to that now, TY for the heads up!

I do want to clarify as to the method I take when I approach tension problems. First I set T = mg (if tension is in the direction opposite of gravity), and from their assess any additional forces. If the block is accelerating downward at a rate less than gravity, then you know that tension PLUS an additional force must be working (MA). If it accelerating downward greater than gravity, you know that tension is equal to mg PLUS an additional force ma. If there is a better way to approach these problems, let me know.

That being said, I'm still not 100% clear, but let me state my confusion a different way.

Problem one: m(9.8m^s2) downward, m(5 m^s2) upward. Net acceleration is still downward, but moving downward more slowly.... And i just realized that I am an idiot. The block is accelerating upward, not downward AT A REDUCED PACE. Tension is countering this force perfectly in 266, creating equilibrium. In 264, there's a NET upward acceleration. Ah ha! That means that WITHOUT that m(5m/s^2) the block would be moving at a constant velocity, because T = mg, so mg = mg. Thanks for bearing with me.

As a follow up question, if a block is attached to a string, and that string is falling and is not yet 'taunt' or being stretched, what is the tension in the string is? (My guess: 0 N)

For question 3, set up a vector equation yourself first. Label all quantities: T, m, g, a if applicable. What is a equal to from problem definition? Which way is your positive direction (imaginary reference axis)?

I realized my answer here was not responding to the proper question. I am creating an answer now.
Thank you so much. You've really helped me. 🙂

 

[SKaminski]

For question 3, set up a vector equation yourself first. Label all quantities: T, m, g, a if applicable. What is a equal to from problem definition? Which way is your positive direction (imaginary reference axis)?

Well, I decided to take your advice and uploaded some pictures online, to help me explain where my confusion is coming from.

Stage 1
http://tinypic.com/view.php?pic=20pepn5&s=6
This is stage 1. It is while the mass is hanging from the rope. If you use 10 m/s^2 as gravity, T = 100N. Done.

Stage 2
http://tinypic.com/view.php?pic=2801f7o&s=6
This is stage 2. It is while the mass is in the process of transition from its lower tension position (while it is hanging) to its raised tension position. Please note that in both the Immediately and Later positions, the ROPE IS SLACK. Since the rope is SLACK, I am assuming T = 0.

Stage 3
http://tinypic.com/view.php?pic=358nmyr&s=6
Stage 3 literally takes place a fraction of a second AFTER Stage 2. Imagine that the block rose an additional 1 cm, and now the rope is taunt.

So, to me, it makes sense that the NET FORCE upward is equal to the tension, because the acceleration is zero, and the velocity is also zero. That means that T = ma.

T = ma.
m = 10 kg, a = 2 m/s^2
T = 20N.

Instead, the credited answer has put T = 120N. Is the problem stating that the ACTUAL upwards acceleration is 12 m/s^2? That would make sense, but i see NO reason to include mg, if the only thing that is providing tension is ma. does this make sense?

 

[LoLCareerGoals]

Well, I decided to take your advice and uploaded some pictures online, to help me explain where my confusion is coming from.

Stage 1
http://tinypic.com/view.php?pic=20pepn5&s=6
This is stage 1. It is while the mass is hanging from the rope. If you use 10 m/s^2 as gravity, T = 100N. Done.

This is correct. But you still didn't set up the vector equation, which may mean you are still memorizing canned formulas. Set up an equation like this:
T - mg = ma (sum of forces on the left, ma on the right), since a = 0, then it reduces to T=mg.

Stage 2
http://tinypic.com/view.php?pic=2801f7o&s=6
This is stage 2. It is while the mass is in the process of transition from its lower tension position (while it is hanging) to its raised tension position. Please note that in both the Immediately and Later positions, the ROPE IS SLACK. Since the rope is SLACK, I am assuming T = 0.

First please abandon your "staged" approach. I think it is causing more confusing than it is helping. You are applying an imaginary force to the block, which is not part of the problem. The force is applied to the rope by pulling on it. The rope is never slack.

Stage 3
http://tinypic.com/view.php?pic=358nmyr&s=6
Stage 3 literally takes place a fraction of a second AFTER Stage 2. Imagine that the block rose an additional 1 cm, and now the rope is taunt.

So, to me, it makes sense that the NET FORCE upward is equal to the tension, because the acceleration is zero, and the velocity is also zero. WHAT???That means that T = ma.

T = ma.
m = 10 kg, a = 2 m/s^2
T = 20N.

Instead, the credited answer has put T = 120N. Is the problem stating that the ACTUAL upwards acceleration is 12 m/s^2? That would make sense, but i see NO reason to include mg, if the only thing that is providing tension is ma. does this make sense?

Ok write out the Newton's law please. In all details, stop skipping steps and making intuitive shortcuts, since they aren't working for you.
To write out the sum of forces acting on the object, look at what is connected to it. What do you see? I only see the rope. So there will be T acting up.
Gravity also never goes away. As long as there is planet Earth and this problem takes place on planet Earth there will be a force equal to mg in magnitude acting down. So still all we have is:
T - mg OR mg - T (depending whether you choose up or down as positive). The right side is ma (if up is positive) or -ma (if down is positive) since acceleration is going up.

So then we have:
T - mg = ma -> T = m(g+a) = 120N or
mg - T = -ma -> -T = -m(g+a) -> T = m(g+a) = 120N

Understand that a is object's NET acceleration not some partial acceleration induced by one of the forces acting on the object.

Here is a made up problem to test you:

Object (10kg) hangs on the rope. An upward force of 60N is applied to the object yet the object remains motionless.
What is the tension of the rope T?
What is the force required to accelerate the object upward at 1m/s^2 if the force is applied in the same manner as in q1? What is T then?

 

[MD Odyssey]

Question 1
My first question is a general one, and is from my own imagination.

Let's say I go sky diving, and I hit terminal velocity. Now, I know that gravity exerts a constant acceleration on me at 9.8 m/s^2. I also know that wind resistance is giving an acceleration in the OPPOSITE direction at 9.8m/s^2. This means that I am in equilibrium. Equilibrium equals no change in velocity, and no change in velocity equals no acceleration.

But... do I have an acceleration? Because I do. I have an acceleration of 9.8 m/s^2 downward, and I have an acceleration of 9.8 m/s^2 upward. Is this correct, or in correct?

What you really mean is net acceleration. There is no net acceleration because the algebraic sum of the forces equals zero. You are correct in identifying a net force of zero as an equilibrium condition.

Question 2
I think this one ties into my previous question regarding net acceleration, but is now stated in terms of Force.
EK 1001 Physics Problem, Question 264:
A 25 kg mass is lowered by a rope. If the velocity of the mass is decreasing at a rate of 5 m/s^2, what is the tension in the rope?

The credited answer is 180, because T = mg + ma. This makes sense if you draw the picture. You have mg (25 * 9.8) pulling down, and ma (25 * 5)pulling up, slowing down the decent. Easy peasy!

EK 1001 Physics Problem, Question 266:
A 32 kg mass is raised by a rope. If the velocity of the mass remains constant at 2 m/s, what is the tension in the rope?

The credited answer is 320N. Because T = mg.

This doesn't make sense to me in light of the question 264, and here's why.

In the previous problem, I used the opposing force + gravitational force to determine the tension. T = (25 * 9.8) + (25 * 5). In this problem, I'm not using the opposing force anymore. If I attempted to solve this problem with the same approach as I tried to solve the previous problem, it would be T = (25 * 9.8) + (25 * 9.8)

This is clearly not correct. The only reasonable explanation I can think of would be that we don't use the 'opposing force' at all. This doesn't seem correct though, because it states (in the answer book for question 264) that the answer is T = mg + ma. I was using 5 m/s^2 for my acceleration upward, and 9.8 m/s^2 for my acceleration downward. Total tension came out correctly. Should I have used 15 m/s^2 for my upward acceleration? That would give me an incorrect answer...

So, if a gravitational opposing force is factored into one answer when it is SLOWING DOWN the effect of gravity, why is a gravitational opposing force not factored in when it is NEGATING gravity? I'm so lost.

Seriously, draw a better picture and all of this will become clear.

Question 3
Physics EK 1001, Question 263
A 10 kg mass hangs from a rope. A force is applied to the rope so that the mass is accelerated upward at 2 m/s^2. What is the tension in the rope.

Credited Answer: 120N.

If i were to nail a string to an overhang, and dangle a weight from it, it would have Tension t1. If I were to pick up that weight, and hold it next to the nail, the tension would be zero, because there would be no tension in the string (the string is slack!). If I were to, next, push the string upward so that it had an acceleration of 2 m/s^2 (and the mass weighed 10kg), shouldn't my Force be 2 m/s^2? Am I thinking about this mystical 'force' the wrong way?

This may contribute to my confusion in Question 2.

Yes. Draw a better picture and it will all become clear.

If it helps at all (not trying to brag) I got 4.0's in both of my physics class in college (graduated 2 years ago).

This means nothing. Introductory college physics has become a joke at most institutions. From looking at your questions, it appears that you need to draw better pictures and be rigorous about the sign of your forces and accelerations. Sometimes I wonder whether or not colleges teach vectors anymore. Draw better pictures and keep track of your unit vectors.