# Dynamics Skier on a Slope Problem

• IbrahimA
In summary, a skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7 m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.
IbrahimA

## Homework Statement

A skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7 m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.

## Homework Equations

Fnetx = (-Ff) + Fa + (-Fgx)
Fnety = Fn + (-Fgy)v2^2 = v1^2 + 2ad

## The Attempt at a Solution

I went by creating two equations:

Fnetx = (-Ff) + Fa + (-Fgx)
ma = -Fn(u) + Fa + mg(sin4.7)
m was canceled out on both sides.
Fn(u) = Fa + g(sin4.7) - a
(1) Fn = Fa + g(sin4.7) - a / u

Fnety = Fn + (-Fgy)
ma = Fn + (m)(-g)(cos4.7)
again canceled out m on both sides
(2) Fn = a + gcos4.7

Correction: Equation 2 is Fn = gcos4.7

And so I did substitution and subbed in both equations to solve for acceleration (a) but the problem is that Fa is also unknown:

(9.8)cos4.7 = Fa + (9.8)(sin4.7) - a / 0.11

So I'm wondering if there is a better way to do this problem, because I feel like I'm doing to wrong.

I included a fbd diagram:

Last edited:
In your equation (2), what acceleration does a represent? Acceleration into the snow?

haruspex said:
In your equation (2), what acceleration does a represent? Acceleration into the snow?

Looking back at equation 2, I feel like the net force on the y-axis is 0. Because from my understanding, the skier is only moving on the x-axis being on the incline, where the forces on the y-axis are cancelling each other out.
Correction of equation 2: Fn = gcos4.7
But even then that still leaves two unknowns, both Fa and a.

IbrahimA said:
But even then that still leaves two unknowns, both Fa and a.
And what applied force does Fa represent?

haruspex said:
And what applied force does Fa represent?

Fa is the applied force that the skier is pushing on the ski poles thus moving up the incline.

IbrahimA said:
Fa is the applied force that the skier is pushing on the ski poles thus moving up the incline.
The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

haruspex said:
The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

I have redone my equations, with:
Equation 1 : Fn = a + gsin4.7 / u
Equation 2 : Fn = mgcos4.7

But again same situation, there are still two unknowns being m and a, since I can't cancel out m on the y-axis as shown below:
Fnety = Fn + (-Fgy)
0 = Fn + (m)(-g)(cos47)
Fn = mgcos4.7

haruspex said:
The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

Algebra Question
In the equation as shown below:
(m)(a) = (u)(g)(m)(cos4.7) + (m)(-g)(sin4.7)
Can I cancel out all the m's on both sides and result into this?:
a = (u)(g)(cos4.7)+(-g)(sin4.7)
Or is that not possible?

IbrahimA said:
Algebra Question
In the equation as shown below:
(m)(a) = (u)(g)(m)(cos4.7) + (m)(-g)(sin4.7)
Can I cancel out all the m's on both sides and result into this?:
a = (u)(g)(cos4.7)+(-g)(sin4.7)
Or is that not possible?
Of course you can cancel the m's. All you need to assume is that m is not zero, and you can safely do that.

IbrahimA
haruspex said:
Of course you can cancel the m's. All you need to assume is that m is not zero, and you can safely do that.

Oh okay, now it makes sense. Thank you.

## What is the "Dynamics Skier on a Slope Problem"?

The "Dynamics Skier on a Slope Problem" is a physics problem that involves a skier moving on a slope with a given angle of inclination and coefficient of friction. The goal is to determine the skier's acceleration and velocity at different points on the slope.

## What are the key factors that affect the solution to this problem?

The key factors that affect the solution to this problem are the angle of inclination of the slope, the coefficient of friction between the skier's skis and the slope, the skier's mass, and the force of gravity.

## How is this problem solved?

This problem is solved using Newton's laws of motion and principles of vector addition. The forces acting on the skier, such as gravity and friction, are calculated and used to determine the skier's acceleration and velocity at different points on the slope.

## What are some real-life applications of this problem?

This problem has real-life applications in skiing and snowboarding, as well as other sports or activities that involve an object or person moving on a slope. It can also be used to understand the motion of objects on inclined surfaces, such as ramps or roller coasters.

One common misconception is that the acceleration and velocity of the skier will be constant throughout the entire slope. In reality, these values will change as the slope's angle and the skier's position on the slope change. Another misconception is that the skier's mass does not affect their acceleration, when in fact it does play a role in the calculation.

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