91Ó°ÊÓ

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 \(15.68\;{\rm{m/s}}\).