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Let's solve the problem step by step:

We need to find the skier's final speed at the top of the incline. Here's how we'll approach it:

### Given:
- Mass of the skier, \( m = 79.9 \, \text{kg} \)
- Initial speed, \( v_i = 16.7 \, \text{m/s} \)
- Height of the incline, \( h = 2.50 \, \text{m} \)
- Angle of incline, \( \theta = 35^\circ \)
- Coefficient of kinetic friction, \( \mu_k = 0.31 \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)

### Step 1: Energy Consideration
This problem involves kinetic energy, gravitational potential energy, and work done by friction. The initial kinetic energy will be partially converted into potential energy and some energy will be lost due to friction.

Using the work-energy principle:

\[
KE_i - W_f = KE_f + PE_f
\]

Where:
- \( KE_i = \frac{1}{2} m v_i^2 \) is the initial kinetic energy.
- \( W_f \) is the work done by friction.
- \( KE_f = \frac{1}{2} m v_f^2 \) is the final kinetic energy (at the top of the rise, the skier has a speed \( v_f \)).
- \( PE_f = mgh \) is the potential energy gained by the skier.

### Step 2: Work Done by Friction
The friction force \( f_f \) is given by:

\[
f_f = \mu_k N
\]

Where \( N \) is the normal force. On an inclined plane, \( N = mg \cos \theta \), so:

\[
f_f = \mu_k mg \cos \theta
\]

The work done by friction \( W_f \) over the distance \( d \) (along the incline) is:

\[
W_f = f_f \cdot d = \mu_k mg \cos \theta \cdot d
\]

We can find \( d \) from the height and angle of the incline using trigonometry:

\[
d = \frac{h}{\sin \theta}
\]

### Step 3: Substituting and Solving
Now let's substitute all the known values and solve for \( v_f \).

#### 1. Initial kinetic energy:
\[
KE_i = \frac{1}{2} \cdot 79.9 \cdot (16.7)^2 = 11138.47 \, \text{J}
\]

#### 2. Potential energy at the top:
\[
PE_f = 79.9 \cdot 9.81 \cdot 2.5 = 1957.48 \, \text{J}
\]

#### 3. Work done by friction:
First, find the distance \( d \):

\[
d = \frac{2.5}{\sin 35^\circ} = 4.357 \, \text{m}
\]

Now calculate the friction force:

\[
f_f = 0.31 \cdot 79.9 \cdot 9.81 \cdot \cos 35^\circ = 201.77 \, \text{N}
\]

So the work done by friction is:

\[
W_f = 201.77 \cdot 4.357 = 878.77 \, \text{J}
\]

#### 4. Final kinetic energy:
Using the energy conservation equation:

\[
11138.47 - 878.77 = \frac{1}{2} \cdot 79.9 \cdot v_f^2 + 1957.48
\]

Simplifying:

\[
10259.70 = 39.95 \cdot v_f^2 + 1957.48
\]

\[
10259.70 - 1957.48 = 39.95 \cdot v_f^2
\]

\[
8302.22 = 39.95 \cdot v_f^2
\]

\[
v_f^2 = \frac{8302.22}{39.95} = 207.84
\]

\[
v_f = \sqrt{207.84} \approx 14.42 \, \text{m/s}
\]

### Final Answer:
The skier's final speed at the top is approximately **14.42 m/s**.

Would you like further details on this solution, or have any questions?

### Relative Questions:
1. How does the angle of the incline affect the skier's final speed?
2. What happens to the final speed if the coefficient of friction increases?
3. How would the problem change if there were no friction?
4. Can you calculate the skier's speed halfway up the incline?
5. What if the skier had a different mass? How would it affect the final speed?

### Tip:
When dealing with inclined planes, always break forces into components parallel and perpendicular to the incline to simplify calculations.

A 79.9 kg skier with an initial speed of 16.7 m/s coasts up a 2.50 m high rise. Find her final speed at the top, given the coefficient of kinetic friction is 0.31.

This problem calculates the final speed of a 79.9 kg skier on a 35-degree incline, using energy conservation principles. Coefficient of friction (0.31) and initial speed (16.7 m/s) are provided. Learn how to apply kinetic and potential energy equations, along with friction work, to find the solution.

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