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

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Homework Help Overview

The discussion revolves around determining the stopping distance of a skier moving down a slope with friction, given an initial speed and specific parameters. The problem involves concepts from dynamics and friction on inclined planes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the formula for stopping distance and question the addition of terms related to gravitational and frictional forces. There is an attempt to analyze forces acting on the skier and how they relate to acceleration.

Discussion Status

Some participants have provided insights into the force analysis and the differences in coordinate systems used in examples. There is recognition of the need to understand the underlying principles rather than relying solely on formulas.

Contextual Notes

Participants note that the textbook example used a different scenario, which may have contributed to confusion regarding the application of the formula. There is an emphasis on the importance of free body diagrams in understanding the problem.

Lannie
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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?
 
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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 [itex]gsin \theta~and~\mu _k gcos \theta[/itex] ?
 
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
[tex]N = mgcos \theta[/tex]

Using the formula

[tex]F_{f} = \mu N[/tex]

[tex]F_{f} = \mu_{k} mgcos \theta[/tex]

X- axis

[tex]F_{f} - mgsin \theta = -ma[/tex]

[tex]\mu_{k} mgcos \theta - mgsin \theta = -ma[/tex]

[tex]mgsin \theta - \mu_{k} mgcos \theta = ma[/tex]

[tex]gsin \theta - \mu_{k} gcos \theta = a[/tex]

[tex]g(sin \theta - \mu_{k} cos \theta) = a[/tex]
 
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.
 

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