# Energy Methods + Pulley + Friction, finding final velocity

1. Mar 30, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
The pulley in the figure has radius R=0.160m and a moment of Inertia I_p = 0.560 kg*m^2. The rope does not slip on the pulley rim. As m_4 (4kg block) drops due to gravity, m_2 (2kg block) is dragged up a ramp inclined at theta = 65 degrees. The coefficient of kinetic friction for the block/ramp is 0.100. Use energy methods to calculate the speed of m_2 after m_4 has dropped 5.00 meters. (Hint the normal force on m_2 is m_2gcos(theta) and change in height is not the same for m_2 and m_4 although they are related to each other).

2. Relevant equations

3. The attempt at a solution
So first, decrease in PE is equal to increase in KE.
4*5*9.8 = 196 J.

The frictional work done is 1.96*cos65*5 = 9.8cos(65)
The force parrallel is 19.6sin(65)*5 = 98sin(65)
Final KE of the system = 196-9.8cos(65)-98sin(65)

Final KE's of the syem
M_4 = 1/2*4*v^2
M_2 = 1/2*2*v^2
Pulley = 1/2*I*w^2 ==> using w=v/R ==> 10.938v^2
Total KE = 13.938v^2

So:
13.938v^2 = 196-9.8cos(65)-98sin(65)
Which results in a 2.72 m/s

Should I have used the coefficient of friction instead of doing what I did? :O
Is there anything wrong with this reasoning?

2. Mar 30, 2015

### Staff: Mentor

Where's the figure?

3. Mar 30, 2015

### RJLiberator

Here it is, sorry.

#### Attached Files:

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4. Mar 30, 2015

### Staff: Mentor

To calculate the work done by friction, you need the friction force. And that requires the coefficient of friction.

5. Mar 30, 2015

### RJLiberator

I calculated it here The frictional work done is 1.96*cos65*5 = 9.8cos(65) which is from mass=2 gravity = 9.2 and u_k = 0.1 = 1.96.

With that in play, does the reasoning/logic of this problem hold up?
Namely: 13.938v^2 = 196-9.8cos(65)-98sin(65)

6. Mar 30, 2015

### Staff: Mentor

You'll have an easier time keeping things straight if you viewed it as $\Delta E = \Delta KE + \Delta PE = Work_{friction}$. (Realizing that the work done by friction is negative, of course.)

7. Mar 30, 2015

### RJLiberator

Also, when they state that the change in y is not the same for both of them, I assume this means that M_4 = 5 while m_2 you must use trigonometry as I did with 5cos(65) and 5sin(65). Correct?

8. Mar 30, 2015

### Staff: Mentor

Yes. The change in height is relevant when finding the change in PE. The 5sin(65) is the change in height of m_2.

9. Mar 30, 2015

### RJLiberator

Hm. If I do that I insert the following:

Change in KE = final - initial
Final I found to be 13.938v_f^2
Initially the system is at rest so KEi = 0.

Change in PE
Final = 4*-5*9.8 =-196J

Force due to friction is 1.96*cos(65)*5 = 4.1416

This nets me the answer of 3.78 = v_f :O

I feel like I am doing something wrong with the change in PE ?

10. Mar 30, 2015

### Staff: Mentor

That's the change in PE of the m_4 mass. What about the other mass?

11. Mar 30, 2015

### RJLiberator

It would then be 2*9.81*5sin65 = 17.76
So -196+17.76 = -178.24

Plugging this back into the equation: 14*x^2-178.24=4.1416

I get x = 3.61

12. Mar 30, 2015

### Staff: Mentor

Redo that bit of arithmetic.

13. Mar 30, 2015

### RJLiberator

Oi. You are absolutely right, I had foolishly forgotten about the 5 in 5sin(65).

So 2*9.8*5sin(65) = 88.82.

So -196+88.82 = -107.18

13.938*x^2-107.18=4.1416

= 2.82 m/s which is similar to my original answer of 2.72 :O

Hm, I'm looking at both methods here and wondering why I am getting the difference of .1.

14. Mar 30, 2015

### RJLiberator

Admittedly, the way you suggested to do it was much more organized and to the point. It just made sense.
I would assume that this method would produce the more exact result, being 2.82.

15. Mar 30, 2015

### Staff: Mentor

Good. It also makes it easier to spot and correct errors.

I haven't checked the arithmetic, but often such differences are due to round off errors.

16. Mar 30, 2015

### RJLiberator

Excellent. Thank you for your help this morning. You clarified this question well for me.