Differentiation Application

Any idea how to do the attached question ?  I have left this question in my assignment and I don't have any idea on it ?

Thx
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Simon LeungAsked:
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leflonCommented:
Hi Simon,

this has been a while.
Hope I kind of remember it correctly. So at least get the general way to solve it.

@(a)
The potential energy
m*g*h

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of mass m1, reduced by mass m2, thus
(m1-m2)*g*h

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gets transfomed into kinetic energy of m2,
1/2*m2*v*v

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once m1 hits the ground. So if  
(m1-m2)*g*h = 1/2*m2*v*v

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you should resolve to v and get the speed.

For (b) I'll need to think a little longer.

leflon
aburrCommented:
Assuming m3 started on table (as per pic) it is 8m above top when m1 hit. You have calculated its speed (part a) It continues upward as a projectile with energy 0.5(m3) v^2. It will go up until all that energy is converted to potential.
KE = pe
0,5 (M3)V^2 = (M3)GH
V^2 = 2GH

REMEMBER TO ADD THE INITIAL 8M
phoffric\Commented:
(b) Isn't this a simple pulley so that when one mass goes down a meter, then the other mass goes up a meter? If so, then if one mass goes down 6 meters, then the other mass goes up 6 meters.
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phoffric\Commented:
I know the picture is a little blurry, but based on the OP, I believe that there is only M1 and M2 and there is no m3.

 My initial inclination was just to use string tension as opposed to potential and kinetic energy . But since potential energy is being used in previous posts, I think it is important to realize that at a split moment before the falling mass hits the table , it has its maximum kinetic energy and zero potential energy ( by defining the tabletop to be at zero potential energy ). However the mass that was rising has at that same moment, both potential energy and kinetic energy . So at that moment the tension in the string becomes zero when the falling object hits the table and the string becomes looser . So the falling object speed is computed at table impact , and that is the same speed as the rising object . This rising object now has only one force on it, namely , the force of gravity . And it is moving straight up with that speed . So now you have standard basic equations to compute how much further up that rising object can travel .

 For this problem, no differential equations are needed.
phoffric\Commented:
Any questions from what all of us have posted? If not, post your new attempt at this problem , and we will check it out .

All - please give the author a chance to digest our comments and ask questions or have a chance to post his attempt at this problem.

 You are constrained to use differentiation, let us know. If you have not learned about potential and kinetic energy yet , let us know and we can talk more about F=m*a.
Simon LeungAuthor Commented:
Does my understanding correct ?

Thx
95e65d82-dc09-4701-a5ba-3f75096a1d0c.jpg
phoffric\Commented:
No.

In physics, when writing an answer, I suggest that (1) you always include units, and (2) that you start with basic principles. Looks to me like you are plugging in numbers based on earlier posts. Let's try again starting at the beginning. Since you feel comfortable using potential and kinetic energy, post your understanding of the basic principle of how these energies are related as a system moves from one state to another state. You need to do this for both parts (a) and (b). Then state this principle in an equation form.

By "state", I mean that your initial state, S0, is that shown in your picture where m1 and m2 are both at rest. State 2 is where m2 is 8 m high and m1 has reached its maximum downward speed (i.e., just hitting the table). S0 and S1 are used for solving part (a). After solving part (a), you will need a third state, S3.
Simon LeungAuthor Commented:
By conservation of Energy, loss of PE of M1 = gain of KE of M2

M1gh = 1/2 M2V^2
(8)(9.8)(6) = 1/2 (3)(V)^2

V = 20.4 m/s

Please comment before I proceed to (b),
leflonCommented:
Simon, phoffric,

my bad, sorry. Thinking to 'complicated'.
phoffric's approach of the defined states is much better.

So if I am not wrong again, (a) looks fine.

leflon
phoffric\Commented:
>> By conservation of Energy, loss of PE of M1 = gain of KE of M2
Actually, I was hoping that you would quote from your book the total conservation of energy law, where the total energy of a system does not change as its state changes provided that there is no friction and no external forces pushing or pulling the masses. Your statement should be general. What you did was to make your statement concrete to your specific problem. Let's give it another try. Write down the principle in words, and then write a general formula using state notation (i.e., subscripts to your terms). This general formula should not have any reference to your specific problem.

After doing that, if you now understand all the terms in this general formula, then I suggest that you provide separate lines for each term and set them to the values that come from your specific problem. Then, you are on your way to solving the problem.

Here is a hint: Not only will you have a (m1-m2) factor in your equation, but you will have a (m1+m2) factor as well.

If you type your response directly into the post instead of a picture, that may make our response faster if we happen to be using a little phone. (For exponentiation, you can use v^2 representing v-squared.)
Simon LeungAuthor Commented:
Sorry, can't get what u are talking about...

Any reference link to your idea that relate to my problem.

Thx again.
phoffric\Commented:
http://spiff.rit.edu/classes/phys211/lectures/gpe/gpe_all.html
Look at the bullets at the top of the of page for useful concepts.

What does your book say about total conservation of energy in a conservAtive force field?
phoffric\Commented:
Note: I used Total energy, but the link refers to Mechanical energy.
Simon LeungAuthor Commented:
M1gh + 0 = M2gh + 0.5M2V^2 + 05M1V^2

V= sqrt( (M1gh - M2gh ) / (0.5M1 + 0.5M2)

= 8.44 m/s

Pls advise.

Thx again.
phoffric\Commented:
https://www.youtube.com/watch?v=vIOgL7jmz78
formulas look reasonable (except for unbalanced parenthesis), but will check tomorrow night. The above link should tell you a lot.

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phoffric\Commented:
And here is a different approach in handling two masses on a simple pulley.
https://www.khanacademy.org/science/physics/forces-newtons-laws/treating-systems/v/two-masses-hanging-from-a-pulley
Simon LeungAuthor Commented:
Noted. Thx
phoffric\Commented:
Your formula is good. Don't forget to include units at all times for initial and final results.

For checking the Arithmetic, try using the alternate approach using string tension and acceleration to see if you get the same numbers.

I think you are ready for part b.
Simon LeungAuthor Commented:
Thx
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