What is the force of kinetic friction acting on the skier in this scenario?

In summary, Tony believes that he has correctly answered part A, but has some trouble with part B. He tries to think of a simpler way to solve the problem, but does not succeed.
  • #1
chmilne
10
0
I believe I have part A correct, but I have having some trouble with part B.

A 63-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.6 m/s. After coasting a distance of 1.9 m up the slope, the speed of the skier is 4.4 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?


Part A

Part A

[tex]W_{nc}[/tex]=(1/2[tex]mv_f[/tex]^2)-(1/2[tex]mv_o[/tex]^2) + ([tex]mgh_f[/tex]) – ([tex]mgh_o[/tex])
[tex]W_{nc}[/tex]= ((½ (63kg) [tex](4.4m/s)^2[/tex]) – (½ (63kg) [tex](6.6 m/s)^2[/tex])) + ((63kg) (9.8m/[tex]s^2[/tex])(.802m) + (63kg)(9.8 m/[tex]s^2[/tex]) (0))
[tex]W_{nc}[/tex]= ((½ (63kg) (19.36 [tex]m^2[/tex]/[tex]s^2[/tex])) – ((½ (63kg) (43.56[tex]m^2[/tex]/[tex]s^2[/tex])) + (495.154 J) + 0
[tex]W_{nc}[/tex]= 609.84J - 1372.14J + 495.154J
[tex]W_{nc}[/tex]= -267.146J



Part B

Because the skier is going up hill, I would think that the angle for the formula W = (Fcos[tex]\theta[/tex])s is 15[tex]5^o[/tex]. So,

-267.146J = (Fcos(155))1.9m
-140.60316 = Fcos(155)
-140.60316 = -.9063078F
155.136N = F

This cannot be true because the professor has mentioned that, to 1 significant digit, the answer is 100N. I'm to find the anser to 2 SF, so, 155.136 would not be true.

Thanks,
Tony
 
Last edited:
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  • #2
The force of friction acts along the full displacemnt of 1.9 m, not a componant of this.
 
  • #3
I get it. The angle should be 180, not 155 because the force run along the x-axis and using 155 would not be the entire x axis. This mean that the answer to part b is 140.603N. Correct?
 
  • #4
I worked this problem and obtained 155N. So I assumed that the problem
means a 1.9m "vertical" rise and obtained the desired answer of 100N.
 
  • #5
The problem states the skier is on a hill that is 25 degrees from the horizontal and that the displacement is 1.9m. So, the 1.9m could not be veritcle, it would also be 25 degrees from the horizontal. However, to answer part b, you need to calculate for 180 degrees, not 155 degrees.
 
  • #6
chmilne said:
I get it. The angle should be 180, not 155 because the force run along the x-axis and using 155 would not be the entire x axis. This mean that the answer to part b is 140.603N. Correct?
Correct (but please, drop those insig digits! It's fingernails on the chalkboard to me!), I get 140 N .

Easier way to think of it; the force vector and the displacement vector are parallel , therefore no vector resolution is required.
 
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  • #7
A simple work and energy Q

I_i
..
. .
. .
. .
.
.
..........
 
  • #8
I tried to draw the q but could not succed:))
 
  • #9
Chi Meson said:
Easier way to think of it; the force vector and the displacement vector are parallel , therefore no vector resolution is required.
Another way to think of it is to remember what a dot product is. chmilne, you start off with the formula [tex]W = F\cos{\theta}s[/tex], but a more general form is [tex]W = \vec{F}\cdot \Delta \vec{s}[/tex].

If you think of it in terms of dot products and vector resolution, then you can mentally and graphically verify that you're selecting the correct angle.

(An even more general form would be [tex]W = \int_{x_1}^{x_2} \vec{F} \,d\vec{x}[/tex], but that's hardly needed at elementary mechanics-level physics.)
 

1. What is the relationship between work and energy for a skier?

Work and energy are directly related for a skier. Work is the force applied to move the skier a certain distance, while energy is the ability to do work. As the skier moves down the slope, work is being done on the skier and the skier's energy is changing.

2. How does the skier's potential energy change as they go down the slope?

The skier's potential energy decreases as they go down the slope. This is because potential energy is the energy an object has due to its position, and as the skier moves down the slope, their position changes and their potential energy decreases.

3. What factors affect the skier's kinetic energy?

The skier's kinetic energy is affected by their mass and their velocity. The greater the mass and velocity of the skier, the greater their kinetic energy will be.

4. How does friction affect the work and energy of the skier?

Friction can decrease the work done on the skier and can also decrease their energy. Friction is a force that opposes the motion of the skier and can slow them down, resulting in less work being done on the skier and a decrease in their energy.

5. Is the skier's energy conserved during their descent?

No, the skier's energy is not conserved during their descent. Some of the skier's potential energy is converted into kinetic energy as they move down the slope, but some energy is also lost due to friction and air resistance. This means that the skier's total energy decreases as they ski down the slope.

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