Question

In \(1899,\) Charles \(\mathrm{M}\). "Mile a Minute" Murphy set a record for speed on a bicycle by pedaling for a mile at an average of $62.3 \mathrm{mph}(27.8 \mathrm{m} / \mathrm{s})$ on a 3 -mi track of plywood planks set over railroad ties in the draft of a Long Island Railroad train. In \(1985,\) a record was set for this type of "motor pacing" by Olympic cyclist John Howard who raced at 152.2 mph \((68.04 \mathrm{m} / \mathrm{s})\) in the wake of a race car at Bonneville Salt Flats. The race car had a modified tail assembly designed to reduce the air drag on the cyclist. What was the kinetic energy of the bicycle plus rider in each of these feats? Assume that the mass of bicycle plus rider is \(70.5 \mathrm{kg}\) in each case.

Short answer

Question: Calculate the kinetic energy of the bicycle plus rider for Mile a Minute Murphy and John Howard, given their speeds and the mass of the bicycle plus rider. Answer: The kinetic energy for Mile a Minute Murphy is approximately 27,040.59 Joules and for John Howard is approximately 166,473.32 Joules.

Step-by-step solution

Calculate the Kinetic Energy for Mile a Minute Murphy

We are given Murphy's speed as \(27.8 \ m/s\) and the mass of the bicycle plus rider as \(70.5 \ kg\). We can now use the kinetic energy formula to calculate the kinetic energy: \(K = \frac{1}{2}mv^2\) \(K = \frac{1}{2}(70.5 \ \text{kg})(27.8 \ \text{m/s})^2\) \(K \approx 27{,}040.59 \ \text{Joules}\) The kinetic energy of the bicycle plus rider for Mile a Minute Murphy is approximately \(27{,}040.59 \ \text{Joules}\).

Calculate the Kinetic Energy for John Howard

We are given Howard's speed as \(68.04 \ m/s\) and the mass of the bicycle plus rider as \(70.5 \ kg\). We can now use the kinetic energy formula to calculate the kinetic energy: \(K = \frac{1}{2}mv^2\) \(K = \frac{1}{2}(70.5 \ \text{kg})(68.04 \ \text{m/s})^2\) \(K \approx 166{,}473.32 \ \text{Joules}\) The kinetic energy of the bicycle plus rider for John Howard is approximately \(166{,}473.32 \ \text{Joules}\).