Optimized Kinetic Energy of Skier on Incline & Horizontal Surface

In summary: I would not have come here if I had book answers all along. But thanks for the...In summary, the skier travels 95 meters before coming to rest.
  • #1
mia_material_x1
22
0
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
 
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  • #2
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
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).
 
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  • #4
stockzahn said:
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).
 

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  • #5
stockzahn said:
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!
 
  • #6
then what would b) be?
 
  • #7
First we have to fix a) !
 
  • #8
BvU said:
First we have to fix a) !
if a) is 95m, then what would b) be? :cry::cry:
 
  • #9
Your working shows 101 m for a). What has to be changed to get the book answer ?

Oh, don't cry, please...
 
  • #10
BvU said:
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
 
  • #11
mia_material_x1 said:
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
 
  • #12
BvU said:
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...
 
  • #13
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
stockzahn said:
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
 
  • #15
mia_material_x1 said:
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?
 
  • #16
180°
 
  • #17
stockzahn said:
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°
 
  • #18
mia_material_x1 said:
180°

Okay then, work is the product of force and displacement. What is the direction of the displacement?
 
  • #19
stockzahn said:
Okay then, work is the product of force and displacement. What is the direction of the displacement?
towards positive x, 20° from x ?
 
  • #20
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.
 
  • #21
stockzahn said:
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
 
  • #22
You had the book answers from the beginning. You sure you understand the exercise now ?
 
  • #23
BvU said:
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
 
  • #24
Ah, I misinterpreted the 95 in post #1. (there basically was no question, so I started commenting)
mia_material_x1 said:
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...
 
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Related to Optimized Kinetic Energy of Skier on Incline & Horizontal Surface

1. What is optimized kinetic energy and how does it relate to skiing?

Optimized kinetic energy refers to the maximum amount of energy that a skier can obtain while skiing on an incline or a horizontal surface. It is directly related to the speed and motion of the skier and determines how efficiently they can navigate down the slope.

2. How does the incline of the slope affect the optimized kinetic energy of a skier?

The incline of the slope has a significant impact on the optimized kinetic energy of a skier. The steeper the slope, the more energy the skier will gain due to the force of gravity pulling them downwards. However, a very steep slope can also increase the risk of losing control and getting injured.

3. What role does the skier's body position play in optimizing kinetic energy?

The skier's body position is crucial in optimizing kinetic energy. A crouched and aerodynamic position can help minimize air resistance and increase speed, thus maximizing kinetic energy. On the other hand, an upright and more relaxed position can reduce speed and energy.

4. How does the type and condition of the ski surface affect the optimized kinetic energy of a skier?

The type and condition of the ski surface can have a significant impact on the optimized kinetic energy of a skier. A smooth and well-groomed surface can provide less resistance and allow the skier to maintain higher speeds and energy. In contrast, a rough or icy surface can slow down the skier and decrease their kinetic energy.

5. Is there a way to calculate or measure the optimized kinetic energy of a skier?

Yes, there are formulas and equations that can be used to calculate the optimized kinetic energy of a skier based on factors such as weight, speed, and slope angle. However, it is also affected by external factors such as wind and air resistance, making it challenging to measure accurately in real-world scenarios.

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