Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a distance of 12.4 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0368. Initially, how fast was the skier going

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, = 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 = ×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

= The kinematic friction = 0.0368

Therefore, for conservation of energy, we have;

1/2·m·v² = ×m×g×s

1/2·v² = ×g×s

v² = 2××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.