A Cyclist on a Slope: Energy, Friction, and Power
A student rides a 15 kg bicycle up a straight road inclined at 5° to the horizontal. Her mass is 50 kg. The road is 200 m long along the slope. She pedals at a steady speed of 4 m/s. A constant friction and air-resistance force of 40 N opposes her motion. Take g = 10 m/s², sin 5° ≈ 0.087, cos 5° ≈ 0.996. She wants to know how much useful work she does against gravity, how much is wasted against friction, and what total power her legs must supply. She notices she feels more tired on this slope than on flat ground at the same speed.
1. Calculate the work done against gravity and the work done against friction over the 200 m climb, and the total useful work output.
Read the worked answer
Total mass m = 50 + 15 = 65 kg.
Vertical height gained h = 200 × sin 5° = 200 × 0.087 = 17.4 m.
Work against gravity = mgh = 65 × 10 × 17.4 = 11,310 J.
Work against friction = 40 N × 200 m = 8,000 J.
Total work output = 11,310 + 8,000 = 19,310 J (≈19.3 kJ).
Note: the work against gravity depends only on the vertical rise, not the path length - using the full 200 m with mg would be wrong.
2. Find the total power her legs must supply at 4 m/s, and explain in energy terms why the slope makes her more tired than flat ground at the same speed.
Read the worked answer
Time to cover 200 m at 4 m/s = 200 / 4 = 50 s.
Power = Work / time = 19,310 / 50 ≈ 386 W.
Equivalently, Power = F_total × v, where the force along the slope = mg sin 5° + friction = (65×10×0.087) + 40 = 56.6 + 40 = 96.6 N; 96.6 × 4 ≈ 386 W. Consistent.
On flat ground there is no height gain, so only the 40 N friction acts: power = 40 × 4 = 160 W. The slope adds ~226 W of gravitational work per second, so she must convert chemical energy to potential energy far faster - hence more fatigue at the same speed.