A 70kg bicyclist, initially at rest at the top of a hill, coasts down the hill. During her ride she experiences a 40N drag. The distance and height of the hill are shown. How fast was the bicyclist traveling at the bottom of the hil?
To determine the speed of the bicyclist at the bottom of the hill, we can use the principle of conservation of energy.
Initially, the bicyclist is at rest at the top of the hill, and therefore has no kinetic energy (KE). At the same time, the bicyclist is at a height h, and therefore has potential energy (PE) given by PE = mgh, where m is the mass of the bicyclist (70 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill.
As the bicyclist coasts down the hill, she experiences a drag force of 40N opposing her motion. This drag force does negative work on the bicyclist, converting her potential energy into kinetic energy and dissipating it as heat due to the drag. Therefore, at the bottom of the hill, the potential energy is converted entirely into kinetic energy.
The formula for kinetic energy is KE = (1/2)mv^2, where v is the speed of the bicyclist at the bottom of the hill.
Equating the initial potential energy and final kinetic energy, we have:
mgh = (1/2)mv^2
Canceling out the mass (m) on both sides of the equation, we have:
gh = (1/2)v^2
Rearranging the equation to solve for v, we get:
v^2 = 2gh
Taking the square root of both sides of the equation, we get:
v = sqrt(2gh)
Substituting the given values, with g = 9.8 m/s^2 and h = 50m, we get:
v = sqrt(2 * 9.8 m/s^2 * 50m) = sqrt(980 m^2/s^2) ≈ 31.3 m/s
Therefore, the speed of the bicyclist at the bottom of the hill is approximately 31.3 m/s.