(III) A bicyclist coasts down a 6.0° hill at a steady speed of . Assuming a total mass of 75 kg (bicycle plus rider), what must be the cyclist’s power output to climb the same hill at the same speed?

The power value can be obtained by evaluating the value of the applied force and the magnitude of the object's speed.

Given data:

The total mass of the bicycle and the rider is \(m = 75\;{\rm{kg}}\).

The angle of inclination of the hill is \(\theta = 6^\circ \).

The speed is \(v = 4.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\)

The following is the free body diagram of a bicyclist coasting down the hill.

Here,\(N\)is the normal force,\(f\)is the frictional force, and\(g\)is the acceleration due to gravity.

Apply the equilibrium condition along the x-direction.

\(\begin{aligned}\sum {F_x} &= ma\\mg\sin \theta - f &= 0\\f &= mg\sin \theta \end{aligned}\)… (i)

The following is the free body diagram of a bicyclist climbing up the hill.

Here,\(F\)is the force required to pull the total mass.

Apply the equilibrium condition along the x-direction.

\(\begin{aligned}\sum {F_x} &= ma\\F - mg\sin \theta - f &= 0\\F &= mg\sin \theta + f\end{aligned}\)

Substitute the value of equation (i) in the above equation.

\(\begin{aligned}F &= mg\sin \theta + mg\sin \theta \\F &= 2mg\sin \theta \end{aligned}\)

The expression for the power is given by:

\(\begin{aligned}P &= Fv\\P &= 2mg\sin \theta v\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}P &= 2\left( {75\;{\rm{kg}}} \right)\left( {9.81\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\sin \left( {6^\circ } \right)\left( {4\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\P &= 614.62\;{\rm{W}}\end{aligned}\)

Thus, the power required by the bicyclist is 614.62 W.