# Homework Help: Pulley system and energy conservation?

1. Feb 28, 2017

### Sagrebella

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

I'm considering the the whole system in my equation

KEi+ Ui + Wnc= Kf+Uf

FN= mghcos

Fkinetic friction = (coefficient of kintetic friction)mghcos

W= Fk(displacement)

3. The attempt at a solution

A.
KEi+ Ui = Kf+Uf

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

m= 10.67

B.
FN= mghcos

Fkinetic friction = (coefficient of kintetic friction)mghcos

(.625)80cos(36.87) = 40

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

20m +200 = 12.5m +80

m=16

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

Thank you!

2. Feb 28, 2017

### kuruman

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

3. Feb 28, 2017

### Sagrebella

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?

4. Feb 28, 2017

### kuruman

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.

5. Feb 28, 2017

### Sagrebella

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?

6. Feb 28, 2017

### kuruman

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.

7. Feb 28, 2017

### Sagrebella

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

8. Feb 28, 2017

### kuruman

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?

9. Feb 28, 2017

### Sagrebella

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

10. Feb 28, 2017

### kuruman

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

11. Feb 28, 2017

### Sagrebella

3 meters ?

12. Mar 1, 2017

### kuruman

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.

13. Mar 1, 2017

### AshUchiha

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

14. Mar 1, 2017

### kuruman

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.

Last edited: Mar 1, 2017
15. Mar 1, 2017

### Sagrebella

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

16. Mar 1, 2017

### kuruman

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