3. (No. 495, pp.) A skier with a mass of m = 60 kg starts from rest and slides down a mountain of height h = 14 m. Then, he moves along a horizontal section without working with his poles and stops. Determine the kinetic friction force acting on the skier until he stops on the horizontal section of length l = 40 m, if 20% of the skier's initial energy was used to overcome resistance while moving down the mountain to the horizontal section.

Solution:

1. Initial energy of the skier: The skier starts from rest at a height \( h = 14 \text{ m} \). The initial energy is entirely potential energy:
   \( E_{initial} = E_p = mgh \)
   \( E_{initial} = 60 \text{ kg} \times 10 \text{ m/s}^2 \times 14 \text{ m} = 8400 \text{ J} \)
2. Energy lost due to resistance on the slope: 20% of the initial energy was lost to overcome resistance.
   \( E_{lost} = 0.20 \times E_{initial} \)
   \( E_{lost} = 0.20 \times 8400 \text{ J} = 1680 \text{ J} \)
3. Energy at the beginning of the horizontal section: The energy remaining is converted into kinetic energy at the start of the horizontal section.
   \( E_{kinetic, start} = E_{initial} - E_{lost} \)
   \( E_{kinetic, start} = 8400 \text{ J} - 1680 \text{ J} = 6720 \text{ J} \)
4. Work done by friction on the horizontal section: On the horizontal section, the kinetic energy is dissipated by the friction force \( F_f \) over a distance \( l = 40 \text{ m} \). The work done by friction is negative, as it opposes motion.
   \( W_f = -F_f \times l \)
5. Final kinetic energy: The skier stops, so the final kinetic energy is 0.
   \( E_{kinetic, final} = 0 \text{ J} \)
6. Applying the work-energy theorem to the horizontal section: The work done by friction equals the change in kinetic energy.
   \( W_f = E_{kinetic, final} - E_{kinetic, start} \)
   \( -F_f \times l = 0 - E_{kinetic, start} \)
   \( -F_f \times 40 \text{ m} = -6720 \text{ J} \)
7. Calculate the friction force:
   \( F_f = \frac{6720 \text{ J}}{40 \text{ m}} \)
   \( F_f = 168 \text{ N} \)

Ответ: Сила трения скольжения равна 168 Н.