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 height is 12m and distance is 65m. How fast was the bicyclist traveling at the bottom of the hill
To find the speed of the bicyclist at the bottom of the hill, we need to calculate the total mechanical energy at the top and bottom of the hill and apply the principle of conservation of energy.
1. Calculate the potential energy at the top of the hill:
Potential energy (PE) = mass (m) * gravity (g) * height (h)
PE = 70 kg * 9.8 m/s^2 * 12 m
PE = 8,176 Joules
2. Calculate the work done against drag during the ride:
Work = force (F) * distance (d)
Work = 40 N * 65 m
Work = 2,600 Joules
3. Calculate the mechanical energy at the top of the hill:
Mechanical energy (ME) = PE + Work
ME = 8,176 J + 2,600 J
ME = 10,776 Joules
4. Apply the principle of conservation of energy:
At the bottom of the hill, all potential energy is converted into kinetic energy and work done against drag.
ME = 1/2 * mass * velocity^2 + Work Against Drag
Rearranging the equation:
1/2 * mass * velocity^2 = ME - Work Against Drag
1/2 * 70 kg * velocity^2 = 10,776 J - 2,600 J
35 kg * velocity^2 = 8,176 J
velocity^2 = (8,176 J) / (35 kg)
velocity^2 = 233.6 m^2/s^2
5. Take the square root of both sides to find the velocity:
velocity = sqrt(233.6 m^2/s^2)
velocity ≈ 15.3 m/s
Therefore, the bicyclist was traveling at approximately 15.3 m/s at the bottom of the hill.