Question

If a bicyclist of mass 65 kg (including the bicycle) can coast down a $6.5^{\mathrm{o}}$hill at a steady speed of $6.0\mathrm{km}/\mathrm{h}$because of air resistance, how much force must be applied to climb the hill at the same speed (and the same air resistance)?

Short answer

The magnitude of the force required to climb the hill is 144.22 N.

Step-by-step solution

Step 1. Statement of Newton’s second law of motion

Newton’s second law of motion states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.

Mathematically,

$a\propto \frac{F_{\mathrm{net}}}{m}$

Step 2. Identification of given data

The mass of the cyclist is $m=65\mathrm{kg}$.

The steepness of the hill is $\theta =6.5^{\mathrm{o}}$.

The speed of the cyclist is $v=6\mathrm{km}/\mathrm{h}$.

Step 3. Drawing an FBD for the cyclist coasting downhill 

A free body diagram of the cyclist coasting down the hill is drawn below.

The forces acting on the moving car are:

• Normal force, N exerted upward on the cyclist
• Weight, W of the cyclist acting downward
• Force of friction due to air, f

Step 4. Determination of the resistive force due to air

As the cyclist is coasting down the hill at a constant speed, the acceleration will be zero. The net force in every direction will be zero. The component of weight equal to the air resistance is $mg\sin \theta$.

Using Newton’s second law of motion down the plane for a constant speed, you get:

$\begin{array}{c}f=mg\sin \theta \\ =\left(65\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\sin {6.5}^{\mathrm{o}}\right)\\ =72.11\mathrm{N}\end{array}$

Step 5. Drawing an FBD for the cyclist climbing the hill

A free body diagram of the cyclist climbing up the hill is drawn below.

The forces acting on the moving car are:

• Normal force, N exerted upward on the cyclist
• Weight, W of the cyclist acting downward
• Force of friction due to air, f
• Force $F_{\mathrm{P}}$pushing the cyclist upward

Step 6. Determination of the force required to climb the hill

The force pushing the cyclist up the hill is $F_{\mathrm{P}}$. The net force acting on the cyclist will be zero because the directions of air resistance and weight are downward.

Using Newton’s second law of motion moving up the plane, you get:

$\begin{array}{c}f+mg\sin \theta -F_{\mathrm{P}}=0\\ F_{\mathrm{P}}=f+mg\sin \theta \\ =2f\end{array}$

Substitute the magnitude of the resistive force due to air in the above expression.

$\begin{array}{c}{F}_{\mathrm{P}}=2\left(72.11\mathrm{N}\right)\\ =144.22\mathrm{N}\end{array}$

Thus, the force required to climb the hill is 144.22 N.