# Kinetic energy

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1. Oct 19, 2015

### mia_material_x1

http://imgur.com/luj32IE 1. The problem statement, all variables and given/known data
A skier starts from rest at the top of a smooth incline of height 20 m as in Figure 2. At the bottom of the
incline, the skier encounters a horizontal rough surface where the coecient of kinetic friction between the
skis and the snow is 0.21.
(a) How far does the skier travel on the horizontal surface before coming to rest?
(b) Find the horizontal distance the skier travels before coming to rest if the incline is also rough with a coecient of kinetic friction equal to 0.21.

2. Relevant equations
Vf² = Vi² + 2a(Xf - Xi)
K = MV²/2

3. The attempt at a solution
Yi =H= 20m
θ = 20°
uk = 0.21

a) Vf² = Vi² - 2g(Yf-Yi)
Vf² = 0 - 2g(0-20m)
Vf = 19.8 m/s

Ke2 - Ke1 = A(friction)

Ke2 - Ke1 = 0 - MV²/2 = - MV²/2
A(f) = F×S×cos(180°-20°)
- MVf²/2 = F×S×cos(180°-20°)
- MVf²/2 = (uk)(Mg)S(cos160°)
S = 101m

b) MVf²/2 = (uk)(Mg)S
S= Vf²/2(ukg) = (19.8m/s) / 2(0.21)(9.8m/s²) = 95m

Last edited: Oct 19, 2015
2. Oct 19, 2015

### BvU

And your question is why are these answers not correct ?
And you already have the correct answer ?
And for the a) question it would be 95 m ?
And the incline is 20 m high and the angle wrt horizontal is 20$^\circ$ ?

3. Oct 19, 2015

### stockzahn

Could you please provide the figure or write what value θ represents?

According to the text of the statement, the answer of the first question (a) should be the one you achieved in (b).

4. Oct 19, 2015

### mia_material_x1

#### Attached Files:

• ###### Screen Shot 2015-10-19 at 7.06.54 AM.png
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5. Oct 19, 2015

### mia_material_x1

I inserted a photo, thanks for reply!

6. Oct 19, 2015

### mia_material_x1

then what would b) be?

7. Oct 19, 2015

### BvU

First we have to fix a) !

8. Oct 19, 2015

### mia_material_x1

if a) is 95m, then what would b) be?

9. Oct 19, 2015

### BvU

Your working shows 101 m for a). What has to be changed to get the book answer ?

Oh, don't cry, please...

10. Oct 19, 2015

### mia_material_x1

I'm not sure... :( Can you enlighten me

11. Oct 19, 2015

### BvU

All correct, except ...
this happens in the right part of your picture. The trajectory is horizontal there

12. Oct 19, 2015

### mia_material_x1

I was thinking A(f) = F×S×cos(20°) seems right, but then I get S= - 21m....

13. Oct 19, 2015

### stockzahn

for a): The velocity of skier you calculated is correct (19.8 m/s).

What is the position of the skier, when he reaches this velocity?

14. Oct 19, 2015

### mia_material_x1

tan20°/20m = (55m)i + (-20m)j @ velocity=19.8m/s

15. Oct 19, 2015

### stockzahn

Well, okay, I just wanted to know if he is still on the slope or already on the horizontal part, but I suppose you know, that he is leaving the slope and entering the horizontal part of his track at this point.

On the horizontal part θ = 0°. What is the direction the friction force is pointing in?

16. Oct 19, 2015

### mia_material_x1

180°

17. Oct 19, 2015

### mia_material_x1

180°

18. Oct 19, 2015

### stockzahn

Okay then, work is the product of force and displacement. What is the direction of the displacement?

19. Oct 19, 2015

### mia_material_x1

towards positive x, 20° from x ?

20. Oct 19, 2015

### stockzahn

The skier is on a horizontal plane, so he only can move parallel to it (as well as the direction of the friction force). Towards positive x is correct, think again about θ. Maybe make a drawing of the situation, if you haven't already done that.