Question

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

Answer 1

The speed with which the skier was going is approximately 2.9906 m/s

Step-by-step 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,
`\mu _k` = 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 =
`\mu _k`×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

`\mu _k` = The kinematic friction = 0.0368

Therefore, for conservation of energy, we have;

1/2·m·v² =
`\mu _k`×m×g×s

1/2·v² =
`\mu _k`×g×s

v² = 2×
`\mu _k`×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.