Question

A bicyclist of mass 75 kg (including the bicycle) can coast down a 4.0degrees hill at a steady speed of 12 km/h. Pumping hard, the cyclist can descend the hill at a speed of 32 km/h.

Using the same power, at what speed can the cyclist climb the same hill? Assume the force of friction is proportional to the square of the speed v; that is, Ffr=bv2, where b is a constant.

Answer 1

Solution:

m = mass = 75 kg
g assumed at 9.82 (m/s)/s
12 km/h = 3.333 m/s
32 km/h = 8.889 m/s

Firstly, calculate the value of b , from coast data ;
(v = 3.333 m/s)
Force behind the bike due to gravity = m * g * sine 4°
= 51.3756 Newtons

If at constant speed, this most also = resistance force
So :
51.3756 = b * v ²
Transpose for b :
b = 51.3756 / v ²
b = 4.62473

out of time, will complete tommorow
sorry

Part 2 : Calculate rider power from downhill pump :
Power required for 8.888 m/s = v ² * b
= 3,248.0955 Watts
Gravity assist power = 51.3756 * 8.888
= 456.6724 W
Rider power = 3,248.0955 - 456.6724
= 2,791.4233 W

Part three : Top speed up incline :
You get :
2,791.4233 = ( v ³ * d ) + ( 51.3756 * v )
By trial and error the top speed uphill = 8.0138 m/s
( 28.85 km/h )