A 75 kg road bicycle racer rides down the highway from Pikes peak. During one stretch of the ride, he descends 5 meters in 12.5 seconds, while accelerating from 35 km/hr to 55 km/hr. Assuming rotational kinetic energy and dissipative forces to be negligible, what average power is supplied by the gravitational force during this time?

Question

Quizplus · Accepted Answer

The average power supplied by the gravitational force can be calculated using the work-energy principle. The work done by gravity is equal to the change in kinetic energy plus the change in potential energy. First, convert speeds from km/hr to m/s: $35 \, \text{km/hr} = 9.72 \, \text{m/s}$ and $55 \, \text{km/hr} = 15.28 \, \text{m/s}$. The change in kinetic energy ($\Delta KE$) is $\frac{1}{2} m (v_f^2 - v_i^2)$, where $m = 75 \, \text{kg}$, $v_f = 15.28 \, \text{m/s}$, and $v_i = 9.72 \, \text{m/s}$. The change in potential energy ($\Delta PE$) is $mgh$, where $g = 9.8 \, \text{m/s}^2$ and $h = -5 \, \text{m}$ (negative because the racer is descending). The total work done by gravity is the sum of $\Delta KE$ and $\Delta PE$. Finally, power is work done over time, so divide the total work by the time interval (12.5 s). Calculating these values gives an average power of approximately 294 W.

A 75 Kg Road Bicycle Racer Rides Down the Highway