A $90\,\text{kg}$ individual is cycling up a hill inclined at $30^\circ$ on a $12\,\text{kg}$ bicycle. The hill is quite steep, and the coefficient of static friction is $0.85$. The cyclist ascends $12\,\text{m}$ up the hill and then pauses at the summit. They then start descending from rest and travel $9\,\text{m}$ before firmly applying the brakes, causing the wheels to lock.

Question

A $90\,\text{kg}$ individual is cycling up a hill inclined at $30^\circ$ on a $12\,\text{kg}$ bicycle. The hill is quite steep, and the coefficient of static friction is $0.85$. The cyclist ascends $12\,\text{m}$ up the hill and then pauses at the summit. They then start descending from rest and travel $9\,\text{m}$ before firmly applying the brakes, causing the wheels to lock.(a) Draw or describe a free-body diagram of the cyclist and bicycle while momentarily at rest on the $30^\circ$ incline.
, (b) Calculate the amount of work performed by friction to stop the bicycle completely, given that the coefficient of kinetic friction is $0.7$.
, (c) Using equations, explain and justify why the cyclist would find it harder to travel up the incline compared to coming down the incline.
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Jason · Accepted Answer

The diagram contains the normal reaction $N$ acting perpendicular to the road surface, the weight $mg$ acting vertically downward, and the static–friction force $f_s$ acting up–slope while the rider is at rest on the hill.
$W_f=-2.6\times10^{4}\,\text{J}$
$F_{\text{up}}=mg\sin\theta+f,\;F_{\text{down}}=mg\sin\theta-f$

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