Question

A 75 kg road bicycle racer rides down the highway from Pike’s peak. During one stretch of the ride, he descends 150 m while accelerating from 35 km/hr to 45 km/hr. Treating his kinetic energy as purely translational (i.e., ignoring any rotational part), how much energy was dissipated by friction and air resistance during this descent?

Answer 1

Given:

Mass, m = 75 kg

Change in distance = 150

Acceleration, a1 = 35 kn/h

Acceleration. a2 = 45 km/h

Let's determine how much energy was dissipated by friction and air resistance during this descent.

To find how much energy was dissipated, apply the formula:

`(W\cdot Dg)+(W\cdot D_f)=\Delta KE`

Rewrite the equation for Wdf:

`WD_f=\Delta KE-WD_g`

Where:

`\begin{gathered} WD_g=75*9.8*150=110250\text{ J} \\ \\ \end{gathered}`

Also:

`\begin{gathered} \Delta KE=(1)/(2)*75*(((45)/(3.6))^2+((35)/(3.6))^2) \\ \\ \Delta KE\text{ = }2314.81\text{ J} \end{gathered}`

WHere Wdf is the energy dissipated, we have:

`\begin{gathered} WD_f=\Delta KE-WDg \\ \\ WD_f=2314.81-110250=-107935\approx108\text{ KJ} \end{gathered}`

ANSWER:

108 KJ