Pulley system and energy conservation?

[Sagrebella]

Homework Statement

[/B]
Two blocks are connected by a string that passes over a massless, frictionless pulley, as shown in the figure. Block A, with a mass mA = 4.00 kg, rests on a ramp measuring 3.0 m vertically and 4.0 m horizontally. Block B hangs vertically below the pulley. Note that you can solve this exercise entirely using forces and the constant-acceleration equations, but see if you can apply energy ideas instead. Use g = 10 m/s2. When the system is released from rest, block A accelerates up the slope and block B accelerates straight down. When block B has fallen through a height h = 2.0 m, its speed is v = 5.00 m/s.

a) Assuming that no friction is acting on block A, what is the mass of block B?


(b) If, instead, there is friction acting on block A, with a coefficient of kinetic friction of 5/8, what is the mass of block B?

Homework Equations

I'm considering the the whole system in my equation[/B]

KE_i+ U_i + W_nc= K_f+U_f

F_N= mghcos

F_{kinetic friction} = (coefficient of kintetic friction)mghcos

W= F_k(displacement)

The Attempt at a Solution

[/B]
A.
KE_i+ U_i = K_f+U_f

0 + m(10)(2) = 0.5m(5)²+ 4(10)(2)

m= 10.67B.
F_N= mghcos

F_{kinetic friction} = (coefficient of kintetic friction)mghcos

(.625)80cos(36.87) = 40

W= F_k(displacement) = 40(5) = 200

20m +200 = 12.5m +80

m=16I included a picture below if my work and equations are not clear. Sorry, I'm not very good at inserting symbols.

Thank you!

 

[kuruman]

KEi+ Ui = Kf+Uf

0 + m(10)(2) = 0.5m(5)2+ 4(10)(

Part (a)
What are the final kinetic and potential energy expressions? Remember that two blocks are moving.

 

[Sagrebella]

Part (a)
What are the final kinetic and potential energy expressions? Remember that two blocks are moving.

I thought that I should treat the blocks as one system: When Block B loses potential energy, Block A gains potential energy. So, Block B has potential energy and final kinetic energy, but Block A only gains final potential energy. Is this not the correct interpretation? How should I write this equation out differently?

 

[kuruman]

... but Block A only gains final potential energy

You need to rethink this. If block A is gaining potential energy, it must be moving and if it is moving, it must have kinetic energy.

 

[Sagrebella]

You need to rethink this. If block A is gaining potential energy, it must be moving and if it is moving, it must have kinetic energy.

ok, so would I have terms for each block i.e. kinetic energy initial for Block B and A, kinetic energy final for B and A, potential energy initial for block A and B, potential energy final for block A and B?

 

[kuruman]

ok, so would I have terms for each block i.e. kinetic energy initial for Block B and A, kinetic energy final for B and A, potential energy initial for block A and B, potential energy final for block A and B

Yes. Also be careful with the potential energy gain of block A. When block B goes down 2 m, block A does not go up 2 m because it is on the incline.

 

[Sagrebella]

Yes. Also be careful with the potential energy gain of block A. When block B goes down 2 m, block A does not go up 2 m because it is on the incline.

Ok. And how would I write the height change for block A. Would it simply be the hypotenuse of the incline? 5 ?

 

[kuruman]

Ok. And how would I write the height change for block A. Would it simply be the hypotenuse of the incline? 5 ?

The incline is a 3-4-5 triangle. For every 5 m the block goes along the incline it moves 4 m horizontally and 3 m vertically. Which of the three displacements is relevant to change in potential energy?

 

[Sagrebella]

The incline is a 3-4-5 triangle. For every 5 m the block goes along the incline it moves 4 m horizontally and 3 m vertically. Which of the three displacements is relevant to change in potential energy?

wouldn't it be the vertical height of 3 m?

 

[kuruman]

It is the vertical height. So when block B drops by 2 m, how high vertically does block A rise?

 

[Sagrebella]

It is the vertical height. So when block B drops by 2 m, how high vertically does block A rise?

3 meters ?

 

[kuruman]

3 meters ?

No. How many meters along the incline does block A move if block B moves down 2 meters? Assume that the string does not stretch or shrink.

 

[AshUchiha]

I'm sorry, how did you solve the B) part?

 

[kuruman]

I'm sorry, how did you solve the B) part?

If you are asking me, forum rules prohibit me from telling you. If you are asking the OP, wait until OP answers (a) first because part (b) is answered along the same lines.

 

[Sagrebella]

If you are asking me, forum rules prohibit me from telling you. If you are asking the OP, wait until OP answers (a) first because part (b) is answered along the same lines.

Yes, it's 2 meters. Now I understand. Actually, I think I figured it out. would you mind taking a look?

 

[kuruman]

Yes, it's 2 meters. Now I understand. Actually, I think I figured it out. would you mind taking a loo

Part (a) is correct. Now onto part (b). The expression for W_NC is incorrect. Use W_NC = f_k s cosθ. Your value for f_k is correct. What is s and what is cosθ?

 

[cjm]

How do we know the angle to use here without it being given?

 

[berkeman]

How do we know the angle to use here without it being given?

Well, we've learned a lot about pulley systems in the last 6 years...

 

[gneill]

Take a look at the image supplied in the first post. You have the height and width of the triangle. Solving for θ should be easy. I included θ below:

 

[cjm]

Part (a) is correct. Now onto part (b). The expression for W_NC is incorrect. Use W_NC = f_k s cosθ. Your value for f_k is correct. What is s and what is cosθ?

If the displacement is 2 meters and the force of friction is the coefficient times normal force, how do we find normal force? I started by finding the y component of hte force of gravity going down. If the mass of block A is 3kg and the g=10, Fg=30N. Then I found used trig to find the y component of Fg where by cos(30)x30=25.98N. Shouldn't normal force be equal in magnitude to the y-component of Fg? If so, I used that value as my Fn and multiplied that by the coefficient given in the question stem as well as the displacement and still got it wrong. What am I missing?

 

[kuruman]

What am I missing?

Sorry, I cannot help you there. To diagnose what you are missing, I will have to know what you have done but you are making that difficult. For example, in post #15 in the bottom half of the page you find a correct expression for the work done by friction. So far so good. Then this equation containing a bunch of numbers without units pops up

What in Newton's name is this and where did it come from?

When you ask for help, you have to make it easy for us to help you. Write out the equations that you are using in symbolic form, solve for what you need to solve also in symbolic form, then substitute the numbers at the very end. That will make your logic and method transparent and make it easy for us to pinpoint where you went wrong.

 

[cjm]

Sorry, I cannot help you there. To diagnose what you are missing, I will have to know what you have done but you are making that difficult. For example, in post #15 in the bottom half of the page you find a correct expression for the work done by friction. So far so good. Then this equation containing a bunch of numbers without units pops up
View attachment 334252
What in Newton's name is this and where did it come from?

When you ask for help, you have to make it easy for us to help you. Write out the equations that you are using in symbolic form, solve for what you need to solve also in symbolic form, then substitute the numbers at the very end. That will make your logic and method transparent and make it easy for us to pinpoint where you went wrong

The image you are referencing is not my own - I am simply trying to learn from the original post. Thanks anyways though!

 

[kuruman]

The image you are referencing is not my own - I am simply trying to learn from the original post. Thanks anyways though!

Yes, I didn't notice that. Sorry for the false alarm. The original post is not the best way to learn from. You should supply your own solution attempt in full so that we have something to look at. Ideally, you should post your equations in LaTeX. Click "LaTeX Guide", lower left t learn how. It's easy to learn and very useful.

Your description of what you did with words is insufficient. You need to show how you implemented the work-energy theorem which is probably where you went wrong.