What is her speed at the bottom of the slope?
A cyclist is coasting at a speed of 11 m/s when she begins to descend a slope that is 410 meters long and 30 meters high. The combined mass of the cyclist and her bicycle is 70 kg. Throughout the descent, a steady drag force of 15 N due to air resistance acts on her. What is her speed at the bottom of the slope?
2 Answers
Θ = arcsin(30/410) = 4.2º
weight downslope = mgsinΘ = 70kg 9.8m/s² sin4.2 = 50.2 N
friction upslope = 15 N
net downslope force = 35.2 N, so
net downslope acceleration = 35.2N / 70kg = 0.50 m/s²
Now use Torricelli's equation:
v² = u² + 2as = (11m/s)² + 2 0.50m/s² 410m = 533 m²/s²
v = 23 m/s
Hope this helps!
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