# Energy Lost by Friction

1. Oct 14, 2009

### Sw0rDz

1. The problem statement, all variables and given/known data
A 22.9-kg girl slides down a playground slide with a vertical drop of 3.29 m. When she reaches the bottom of the slide, her speed is 1.34 m/s. How much energy was dissipated by friction?

2. Relevant equations
I used $$\Delta$$KE + $$\Delta$$PE = 0

3. The attempt at a solution
I got KEinitial = 0
KEfinal = 1/2 * 22.9kg * (1.34)2 = 20.56

I got PEinititial = m * g * h = 22.9 * 9.8 * 3.29 = 738.34
PEfinal = 0

20.56 (Friction) = 738.34

Is that on the right track? Where would I go from there?

2. Oct 14, 2009

### gabbagabbahey

Hi Sw0rDz, welcome to PF!

If you are using PE to represent gravitational potential energy, then there is another term that should be included in this equation; the work done by friction Wf:

$$\Delta\text{KE}+\Delta\text{PE}+W_f=0$$

make sense?

P.S. In the future, problem like this should go in the Introductory physics folder

3. Oct 14, 2009

### Staff: Mentor

This would be true if mechanical energy was conserved--there was no friction--but that's not the case here. How would you modify this to include energy dissipated by friction?

This looks good.

I don't know where you got this or what it means.

Hint: Compare the initial mechanical energy with the final.

4. Oct 14, 2009

### Sw0rDz

Thanks! That is what I needed.

Second part of the problem is:
If the slide is inclined at 20.5° with the horizontal, what is the coefficient of kinetic friction between the girl and the slide?

KE doesn't depend on the angle, so I left it the same.

I took PE to be m * g * cos(20.5) *h

Is that right?

5. Oct 14, 2009

### gabbagabbahey

No, neither energy term will depend on the angle....what will depend on the angle are the components of the gravitational and frictional forces parallel and normal to the slide.

6. Oct 15, 2009

### Sw0rDz

Energy lost due to work will be the same. Just when trying to calculate the coefficient of friction will depend on the angle?

7. Oct 15, 2009

### gabbagabbahey

What direction does the force of friction act in? What is the formula for finding its magnitude?

8. Oct 15, 2009

### Sw0rDz

In my diagram, have it acting left (negative)

Previously I got.
Wfriction = 717.79J
(Was right according to LON-CAPA and previous equation)

Now I set. 717.79 = $$\upsilon$$KF * M * G * Cos($$\Theta$$)

9. Oct 15, 2009

### gabbagabbahey

Where did you get that formula from?

What is the general formula for frictional force? What is the general formula (involves an integral) for work done by a force?

10. Oct 15, 2009

### Sw0rDz

Ffriction = (Coefficient of KF) * (Force)
W = $$\int$$ F * dl

Where dl = infinitive displacement.

11. Oct 15, 2009

### gabbagabbahey

Right, and what is the normal force in this case?

12. Oct 15, 2009

### Sw0rDz

$$F_n_o_r_m_a_l = M *G * Cos (\theta)$$

13. Oct 15, 2009

### gabbagabbahey

You should probably use a lowercase 'g' so as not to confuse it with the universal gravitational constant, but yes.

So, $F_f=\mu_f mg\cos\theta$...what about its direction?

14. Oct 15, 2009

### Sw0rDz

Negative because it acts against the positive kinetic energy.

Maybe..
$$-F_f=\mu_f mg\cos\theta$$

15. Oct 15, 2009

### gabbagabbahey

"Negative" isn't a direction, it's just a sign... does the frictional force act to the left? to the right? up? down? tangent to the slide? normal to the slide?

16. Oct 15, 2009

### Sw0rDz

In my diagram. Positive direction is to the right. Negative is to the left. Friction goes to the left.
Tangent to the slide, I think..

Last edited: Oct 15, 2009
17. Oct 15, 2009

### gabbagabbahey

Directly to the left? Shouldn't it be going tangent to the slide, opposite to the girl's velocity?

18. Oct 15, 2009

### Sw0rDz

Yes because friction is based of the Normal Force. And Normal force is perpendicular to the tangent line of the slide.

19. Oct 15, 2009

### gabbagabbahey

Okay, so wha soes that tell you about $\textbf{F}_f\cdot d\textbf{l}$? And what do you get when you integrate it over the length of the slide? How long is the slide?

20. Oct 15, 2009

### Sw0rDz

$$\int_0^{3.2} \left( ce_{KF} \cdot m\cdot g\cdot cos(\theta)\cdot h\right) dh$$

$$\left[ ce_{KF} \cdot m\cdotg\cdot g\cdot cos(\theta) \cdot \frac{1}{2} h^{2} \right]_{0}^{3.2}$$

??

Last edited: Oct 15, 2009