What Energy transformations when climbing a hill?
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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, = 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.