A 67.6-kg skier coasts up a snow-covered hill that makes an angle of 21.8 degrees with the horizontal. The initial speed of the skier is 9.75 m/s. After coasting a distance of 1.02 m up the slope, the speed of the skier is 4.97 m/s. Find the work done by the kinetic frictional force that acts on the skis. What is the magnitude of the kinetic frictional force?

To solve this problem, we will use the work-energy principle and the concept of forces acting on the skier.

### Given Data:
- Mass of the skier, \( m = 67.6 \, \text{kg} \)
- Initial speed, \( v_i = 9.75 \, \text{m/s} \)
- Final speed, \( v_f = 4.97 \, \text{m/s} \)
- Distance up the slope, \( d = 1.02 \, \text{m} \)
- Angle of the hill, \( \theta = 21.8^\circ \)

### Steps to Find the Work Done by Kinetic Friction:

1. **Calculate the initial and final kinetic energy (KE)**:
   \[
   \text{KE}_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 67.6 \times (9.75)^2
   \]
   \[
   \text{KE}_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 67.6 \times (4.97)^2
   \]

2. **Calculate the change in kinetic energy**:
   \[
   \Delta \text{KE} = \text{KE}_f - \text{KE}_i
   \]

3. **Calculate the gravitational potential energy (PE) gained while moving up the slope**:
   \[
   \text{PE} = mgh = mgd \sin(\theta)
   \]
   where \( h = d \sin(\theta) \).

4. **Calculate the work done by kinetic friction**:
   The work done by friction can be expressed as:
   \[
   W_f = \Delta \text{KE} + \text{PE}
   \]

5. **Find the magnitude of the frictional force**:
   The work done by friction can also be expressed as:
   \[
   W_f = -F_f \cdot d
   \]
   Rearranging gives:
   \[
   F_f = -\frac{W_f}{d}
   \]

### Calculation:

1. **Initial kinetic energy**:
   \[
   \text{KE}_i = \frac{1}{2} \times 67.6 \times (9.75)^2 = 3226.66 \, \text{J}
   \]

2. **Final kinetic energy**:
   \[
   \text{KE}_f = \frac{1}{2} \times 67.6 \times (4.97)^2 = 835.59 \, \text{J}
   \]

3. **Change in kinetic energy**:
   \[
   \Delta \text{KE} = 835.59 - 3226.66 = -2391.07 \, \text{J}
   \]

4. **Gravitational potential energy**:
   \[
   h = d \sin(\theta) = 1.02 \times \sin(21.8^\circ) = 1.02 \times 0.370 = 0.3774 \, \text{m}
   \]
   \[
   \text{PE} = mgh = 67.6 \times 9.81 \times 0.3774 = 249.29 \, \text{J}
   \]

5. **Work done by friction**:
   \[
   W_f = \Delta \text{KE} + \text{PE} = -2391.07 + 249.29 = -2141.78 \, \text{J}
   \]

6. **Magnitude of the frictional force**:
   \[
   F_f = -\frac{W_f}{d} = -\frac{-2141.78}{1.02} = 2093.36 \, \text{N}
   \]

### Final Answers:
- The work done by the kinetic frictional force is approximately **-2141.78 J**.
- The magnitude of the kinetic frictional force is approximately **2093.36 N**.

If you have any questions or need further details, feel free to ask!

### Further Questions:
1. How does the angle of the slope affect the work done by friction?
2. What role does the mass of the skier play in the calculations?
3. How would the work done change if the skier had a different initial speed?
4. Can you explain how gravitational potential energy is calculated in this scenario?
5. What would happen if the slope was frictionless?

**Tip:** Always break down the problem into manageable parts and use known physics formulas to approach complex scenarios systematically.

Explore how to calculate the work done by kinetic friction on a skier coasting up a hill. Given a mass of 67.6 kg, initial speed of 9.75 m/s, and a final speed of 4.97 m/s after 1.02 m, find the kinetic friction force and work done with detailed calculations.

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