What Is the Correct Formula for a Skier's Stopping Distance on a Slope?

AI Thread Summary
The discussion focuses on calculating the stopping distance of a skier moving down a slope with friction, using an initial speed of 22.2 m/s, a coefficient of kinetic friction (μk) of 0.173, and a slope angle (θ) of 4.9 degrees. The initial formula attempted was Xinitial = Vsquared/2g(sinθ + μkcosθ), which led to an incorrect stopping distance calculation. Participants clarify that the issue arises from misunderstanding the forces acting on the skier, emphasizing the need to analyze forces using free body diagrams. The correct approach involves recognizing that the textbook example involved a different scenario, specifically a block sliding up the slope rather than down. Understanding the forces and their directions is crucial for accurately determining stopping distances in such physics problems.
Lannie
Messages
11
Reaction score
0
I'm trying to work out a problem involving the stopping distance of a skiier, and I thought it was a straightforward problem but I've gone wrong somewhere.

Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 22.2 m/s.
Assume μk=0.173 and θ=4.9

To solve, I tried the formula:

Xinitial= Vsquared/2g(sinθ + μkcosθ)
so i used the data given and got

X=(22.2)squared/2(9.8)(sin4.9+0.173cos4.9)
= 97.5 m

but this did not work out.

Anyone have any thoughts?
 
Physics news on Phys.org
Lannie said:
I'm trying to work out a problem involving the stopping distance of a skiier, and I thought it was a straightforward problem but I've gone wrong somewhere.

Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 22.2 m/s.
Assume μk=0.173 and θ=4.9

To solve, I tried the formula:

Xinitial= Vsquared/2g(sinθ + μkcosθ)

Here's your problem : why are you adding gsin \theta~and~\mu _k gcos \theta ?
 
that was the formula given in the textbook in the section on stopping distances.

based on X1= - Vsquared/2a
(all in the x direction)

and 2a= 2g x (sinθ + μkcosθ)

i'm ignorant-- why doesn't that work?
 
Well doing the force analysis

Y-axis
N = mgcos \theta

Using the formula

F_{f} = \mu N

F_{f} = \mu_{k} mgcos \theta

X- axis

F_{f} - mgsin \theta = -ma

\mu_{k} mgcos \theta - mgsin \theta = -ma

mgsin \theta - \mu_{k} mgcos \theta = ma

gsin \theta - \mu_{k} gcos \theta = a

g(sin \theta - \mu_{k} cos \theta) = a
 
ok that definitely makes sense.
the example in the textbook had a different co-ordinate system so I guess that was the difference.
thank you both for the help!
 
Lannie, instead of trying to use formulas blindly, learn how to figure them out using free body diagrams.

And the example in your textbook had the block slidingup the slope, not down it. That was the difference, not the co-ordinate system.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top