Math Problem Statement

A skier is gliding along at 3.2 m/s on horizontal, frictionless snow. He suddenly starts down a 10∘ incline. His speed at the bottom is 20 m/s . What is the length of the incline?

Solution

To solve for the length of the incline, we can use the concept of energy conservation, as there is no friction involved and only gravitational potential energy is being converted into kinetic energy.

Given dataTo solve for the length of the incline, we can use the concept of energy conservation, as there is no friction involved and only gravitational potential energy is being converted into kinetic energy.

Given

  • Initial speed vi=3.2m/sv_i = 3.2 \, \text{m/s}
  • Final speed vf=20m/sv_f = 20 \, \text{m/s}
  • Incline angle θ=10\theta = 10^\circ

We will use the principle of conservation of mechanical energy. The skier's initial energy consists of kinetic energy at the top of the incline, and at the bottom, it's fully converted to kinetic energy.

Step 1: Kinetic Energy at the Top and Bottom

The skier has only kinetic energy both at the top and the bottom. Since the snow is frictionless, potential energy is lost as the skier descends the incline and converted into kinetic energy.

  • Initial kinetic energy: KEi=12mvi2KE_i = \frac{1}{2} m v_i^2
  • Final kinetic energy: KEf=12mvf2KE_f = \frac{1}{2} m v_f^2

Step 2: Energy Conservation

The difference in kinetic energy is equal to the loss in potential energy:

KEfKEi=mghKE_f - KE_i = mgh

Where hh is the vertical height of the incline and can be related to the incline's length LL and angle θ\theta as:

h=Lsinθh = L \sin \theta

Thus, the equation becomes:

12mvf212mvi2=mgLsinθ\frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = mgL \sin \theta

Step 3: Simplifying and Solving for LL

The mass mm cancels out from all terms, and we can solve for the length LL:

12(vf2vi2)=gLsinθ\frac{1}{2} (v_f^2 - v_i^2) = g L \sin \theta

Rearranging for LL:

L=vf2vi22gsinθL = \frac{v_f^2 - v_i^2}{2g \sin \theta}

Step 4: Substituting the Values

  • vf=20m/sv_f = 20 \, \text{m/s}
  • vi=3.2m/sv_i = 3.2 \, \text{m/s}
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2
  • θ=10\theta = 10^\circ

Now, calculate LL:

L=(20)2(3.2)229.8sin(10)L = \frac{(20)^2 - (3.2)^2}{2 \cdot 9.8 \cdot \sin(10^\circ)}

Let me compute the result.The length of the incline is approximately 114.52meters114.52 \, \text{meters}.

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:

  1. How would the result change if the angle of the incline increased to 1515^\circ?
  2. What would happen if friction were introduced? How would it affect the length of the incline?
  3. How can we calculate the skier's time to reach the bottom of the incline?
  4. What is the skier's acceleration while descending the incline?
  5. How would this scenario differ if the skier started from rest at the top?

Tip: When solving incline problems, always relate height to the length of the incline using trigonometric functions like sine or cosine, based on the problem's geometry.

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Math Problem Analysis

Mathematical Concepts

Energy Conservation
Kinematics
Trigonometry

Formulas

Kinetic Energy: KE = (1/2)mv^2
Conservation of Energy: KE_f - KE_i = mgh
Height of incline: h = L sin(θ)
Length of incline: L = (v_f^2 - v_i^2) / (2g sin(θ))

Theorems

Conservation of Mechanical Energy

Suitable Grade Level

Grades 10-12