Optimized Kinetic Energy of Skier on Incline & Horizontal Surface

[mia_material_x1]

http://imgur.com/luj32IE 1. Homework Statement
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.

Homework Equations

Vf² = Vi² + 2a(Xf - Xi)
K = MV²/2

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 = 101mb) MVf²/2 = (uk)(Mg)S
S= Vf²/2(ukg) = (19.8m/s) / 2(0.21)(9.8m/s²) = 95m

 

[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$^\circ$ ?

 

[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).

 

[mia_material_x1]

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).

 

[mia_material_x1]

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).

I inserted a photo, thanks for reply!

 

[mia_material_x1]

then what would b) be?

 

[BvU]

First we have to fix a) !

 

[mia_material_x1]

First we have to fix a) !

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

 

[BvU]

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

Oh, don't cry, please...

 

[mia_material_x1]

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

Oh, don't cry, please...

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

 

[BvU]

Ke2 - Ke1 = A(friction)

Ke2 - Ke1 = 0 - MV²/2 = - MV²/2
A(f) = F×S×cos(180°-20°)

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

 

[mia_material_x1]

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

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

 

[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?

 

[mia_material_x1]

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?

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

 

[stockzahn]

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

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?

 

[mia_material_x1]

180°

 

[mia_material_x1]

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?

180°

 

[stockzahn]

180°

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

 

[mia_material_x1]

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

towards positive x, 20° from x ?

 

[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.

 

[mia_material_x1]

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.

mgH = umgx+umgd
H=ux +ud
d= (H-ux)/u = [20m-(0.21)(55m)]/0.21 = 40m
got it, thank youuu xD

 

[BvU]

You had the book answers from the beginning. You sure you understand the exercise now ?

 

[mia_material_x1]

You had the book answers from the beginning. You sure you understand the exercise now ?

I would not have come here if I had book answers all along. But thanks for the help

 

[BvU]

Ah, I misinterpreted the 95 in post #1. (there basically was no question, so I started commenting)

I would not have come here if I had book answers all along. But thanks for the help

You would be surprised to see how many posters do have the solution manual and still run into problems (It happened to me a lot as well).

[edit] never mind the FBD. I made a few calclation errors as well. Ahem...