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

In summary, the conversation discusses a problem involving the stopping distance of a skier moving down a slope with friction. The formula Xinitial= Vsquared/2g(sinθ + μkcosθ) is used to solve the problem, but it does not work out. After further analysis and clarification, it is determined that the example in the textbook had a different coordinate system and direction of movement, leading to the discrepancy. The conversation ends with a suggestion to learn how to figure out formulas using free body diagrams.
  • #1
Lannie
11
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?
 
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  • #2
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] ?
 
  • #3
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?
 
  • #4
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]
 
  • #5
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!
 
  • #6
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.
 

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

What is friction?

Friction is a force that opposes motion between two surfaces that are in contact with each other. It is caused by the microscopic roughness of the two surfaces and the interlocking of the irregularities.

How does friction affect stopping distance?

Friction plays a crucial role in stopping distance as it helps to slow down or stop an object in motion. The more friction between the object and the surface it is moving on, the shorter the stopping distance will be.

What factors affect the amount of friction between two surfaces?

The amount of friction between two surfaces is affected by factors such as the nature of the surfaces (smooth or rough), the force pressing the surfaces together, and the type of motion (sliding, rolling, or static).

How can friction be reduced to decrease stopping distance?

Friction can be reduced by using materials with smoother surfaces, such as lubricants or slippery coatings. Additionally, reducing the force pressing the surfaces together and minimizing sliding or rolling motion can also decrease friction and stopping distance.

What are some real-world examples of friction and stopping distance?

Some real-world examples of friction and stopping distance include a car's brakes rubbing against the wheels to slow down or stop the car, a person's shoes gripping the ground to prevent slipping and falling, and a ball rolling to a stop on a rough surface due to the friction between the ball and the ground.

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