A cyclist starts from rest and coasts down a hill. The mass of the cyclist plus bicycle is 85 kg. After the cyclist has traveled 180 m, (a) what was the net work done by gravity on the cyclist? (b) How fast is the cyclist going? Ignore air resistance and friction.

The work done depends on the force exerted on a body and the displacement of the body. It is given by:

\(\begin{aligned}W &= \vec F \cdot \vec d\\ &= Fd\cos \theta \end{aligned}\)

Here, \(\theta \) is the angle between the force and the displacement vector.

The mass of the cyclist is\(m = 85\;{\rm{kg}}\).

The distance traveled by the cyclist is\(d = 180\;{\rm{m}}\).

The steepness of the hill is \(\theta = {4^{\rm{o}}}\).

The angle between the weight (force of gravity) and the velocity of the cyclist is:

\(\begin{aligned}\theta ' &= {90^{\rm{o}}} - \theta \\ &= {90^{\rm{o}}} - {4^{\rm{o}}}\\ &= {86^{\rm{o}}}\end{aligned}\)

The force of gravity or the weight of the cyclist is given by:

\({F_{\rm{g}}} = mg\)

The work done by the force of gravity will be:

\(\begin{aligned}{W_{\rm{g}}} &= {F_{\rm{g}}}d\cos \theta \\ &= mgd\cos \theta \\ &= \left( {85\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/s}}} \right)\left( {180\;{\rm{m}}} \right)\cos \left( {{{86}^{\rm{o}}}} \right)\\ &\approx 10460\;{\rm{J}}\end{aligned}\)

Thus, the work done by the gravitational force on the cyclist is 10460 J.

According to the work-energy theorem, the change in kinetic energy of the body is equal to the work done by the net force. Mathematically,

\(\begin{aligned}{W_{{\rm{net}}}} &= K{E_{\rm{f}}} - K{E_{\rm{i}}}\\ &= \Delta KE\end{aligned}\)

The kinetic energy of the cyclist after coasting downhill is equal to the work done on the cyclist due to the gravitational force. From the work-energy theorem,

\(\begin{aligned}{W_{\rm{g}}} &= KE\\{W_{\rm{g}}} &= \frac{1}{2}m{v^2}\\{v^2} &= \frac{{2{W_{\rm{g}}}}}{m}\\v &= \sqrt {\frac{{2{W_{\rm{g}}}}}{m}} \end{aligned}\)

Substitute the known numerical values in the above expression for the velocity,

you get:

\(\begin{aligned}v &= \sqrt {\frac{{2\left( {10460\;{\rm{J}}} \right)}}{{85\;{\rm{kg}}}}} \\ &= 15.68\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the cyclist going down the hill is \(15.68\;{\rm{m/s}}\).