Answer:
The speed with which the skier was going is approximately 2.9906 m/s
Explanation:
The given parameters are;
The distance the skier slides before coming to rest, s = 12.4 m
The coefficient of friction between the skier and the snow, [tex]\mu _k[/tex] = 0.0368
Therefore, for conservation of energy, we have;
Initial kinetic energy = Work done by the kinetic friction force
Initial kinetic energy = 1/2·m·v²
The work done by the kinetic friction force = [tex]\mu _k[/tex]×m×g×s
Where;
m = The mass of the skier
v = The speed with which the skier was going
g = The acceleration due to gravity = 9.8 m/s²
s = The distance the skier slides before coming to rest = 12.4 m
[tex]\mu _k[/tex] = The kinematic friction = 0.0368
Therefore, for conservation of energy, we have;
1/2·m·v² = [tex]\mu _k[/tex]×m×g×s
1/2·v² = [tex]\mu _k[/tex]×g×s
v² = 2×[tex]\mu _k[/tex]×g×s = 2 × 0.0368 × 9.8 × 12.4 = 8.943872
v = √(8.943872) ≈ 2.99063070271 ≈ 2.9906 m/s
The speed with which the skier was going = v ≈ 2.9906 m/s.
