A bicyclist starting from rest applies a force of F = 289 N to ride his bicycle

Question

A bicyclist starting from rest applies a force of F = 289 N to ride his bicycle across flat ground for a distance of d = 325 m before encountering a hill making an angle of theta = 22 degrees with respect to the horizontal. The bicycle and rider have a mass of m = 122 kg combined. Part (a) How much work. IF in joules, did the rider do before reaching the lull? Part (b) What is the bicycle's speed, v in m s. just before the hill? Assume there is no kinetic friction between the bicycle and the ground. Part (c) If the cyclist starts coasting at the bottom of the hill, what distance, d_i in meters, does the bike travel up the incline?

Explanation / Answer

You have got the part (a) and (b) correct, so I will directly do (c)

(c)

Before cyclist starts coasting, kinetic energy of the cyclist+cycle is;

KE = F.d = (289 N)(325 m)

or, KE = 93925 J

Let the cylist reaches a height h from the ground on the incline, then he stops, so loss in KE is gain in PE;

So mgh = KE = 93925 J

or, (122 kg)(9.8 m/s2)h = 93925 J

or, h = 78.56 m

So, we have;

sin220 = h/di

or, di = h/sin220 = 78.56 m/sin220

or, di = 209.7 m

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