How Does Friction and Air Resistance Affect a Skier's Speed on a Slope?

[ae4jm]

[SOLVED] Motion on a Plane

Homework Statement

A professional skier's initial acceleration on fresh snow is 90% of the acceleration expected on a frictionless, inclined plane, the loss being due to friction. Due to air resistance, his acceleration slowly decreases as he picks up speed. The speed record on a mountain in Oregon is 180 kilometers per hour at the bottom of a 25 degree slope that drops 200 m.

a) What exit speed could a skier reach in the absence of air resistance?
b) What percentage of this ideal speed is lost to air resistance?

Homework Equations

Vf^2=Vi^2+2at
a= (9.8)sin (theta)

The Attempt at a Solution

The correct answer for part a is 214 km/hr and the correct answer for part b is 16%.

Do I take the 90% of the acceleration after finding acceleration with a= (9.8)sin (theta), I just need to add the 10% back to the acceleration?

The 200m drop; is this the opposite of the angle or is this the hypotenuse? I believe it is the hypotenuse, but it if it was the opposite of the angle I should be able to find the adjacent and hypotenuse of the angle with inverse tangent, correct?

Please, if you have any tips or pointers for this problem I am grateful to hear them!

 

[hage567]

There is an error in your first equation you've stated "Relevant Equations". It should be a "d", not a "t".

Do I take the 90% of the acceleration after finding acceleration with a= (9.8)sin (theta),

Yes. This is the one taking into account the friction with the snow, not the air.

I just need to add the 10% back to the acceleration?

Not sure what you mean by this.

The 200m drop; is this the opposite of the angle or is this the hypotenuse? I believe it is the hypotenuse, but it if it was the opposite of the angle I should be able to find the adjacent and hypotenuse of the angle with inverse tangent, correct?

I would take this to be the vertical distance from the top of the hill to the bottom (so not the hypotenuse). You can use trig to find the length of the hypotenuse.

 

[ae4jm]

The answer I get is not the same as the book's answer.

First,
I solved for the hypotenuse, my s variable, to get the distance traveled by the skier. I go ahead and take .90 of the acceleration to plug into the original information. I use Vf^2=Vi^2+2a(s). I convert the 180km/hr into 50m/s and use Vi=0m/s. So, my first equation used will be, 50^2=0+2(3.96)s. I solved for s and got s=315.66m.

Then I got back and plug in s=315.66 into the equation, Vf^2=Vi^2+2a(s) to find my Vf for the final speed. I also use a=4.4, since I'm not taking 90% of the acceleration anymore. I get Vf^2=0+2(4.4)(315.66). Vf=52.7 m/s. Now I convert from m/s to km/hr and I get 189.72km/hr, which is off from 214 km/hr.

Does anyone see any errors?

 

[ever1809]

Here's how to solve the problem.

For part A:

You know that Vf^2=Vi^2+2a(distance of slope)

Since Vi^2=0, you have Vf^2=2a(distance of slope)
You also want to factor in 90% acceleration.

So, plugging in the variables you have:
Vf^2=(2)(9.8)(200)(.90)
Vf^2=3,528
Vf=(3,528)^(1/2)
Vf=59.39697

Converting into km/hr,
59.39697 m/s * 3600 s/hr * .001 km/m

You get 213.829 km/hr, which rounds to 214 km/hr.


Part B is the easy part.
180/214=84%
So you have lost 100%-84%=16%.

 

[3F1A]

Actually, ever1809 is close but incorrect:

Vf^2=(2)(9.8)(200)(.90) is missing a component and the correct distance.

the component that is affected by gravity: (g*sin(theta))(0.9) = 3.7275 m/s^2
the distance is our hypotenuse = 200m/sin(25) = 473 m

Vf^2 = (2)(3.7275)(473)

ever1809 was correct in the problem's followthru and just by accident the answer comes out the same so this was just an example of a truly poorly planned and misleading problem