Question

(III) A cyclist intends to cycle up a 7.50° hill whose vertical height is 125 m. The pedals turn in a circle of diameter 36.0 cm. Assuming the mass of bicycle plus person is 75.0 kg, (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves the bike 5.10 m along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.

Short answer

(a) The work done against gravity is 91875 J.

(b) The tangential force exerted on the pedals is 432.84 N.

Step-by-step solution

Given data

The work done against gravity is equal to the gain in potential energy.

Given data:

The total mass of the bicycle and person is\(m = 75.0\;{\rm{kg}}\).

The height of the hill is\(h = 125\;{\rm{m}}\).

The angle of incline is \(\theta = {7.50^ \circ }\).

The diameter of the circle where the pedals turn is\(d = 36.0\;{\rm{cm}}\).

The bike moves\(x = 5.10\;{\rm{m}}\)in each revolution of the pedals.

Assumptions:

Let F be the average force on the pedals tangent to their circular path.

Let the cycle goes \(h'\) height during one revolution of the pedals.

Calculation for part (a)

Work done against gravity is equals to the change in potential energy.

Therefore, the work done against gravity is calculated as follows:

\(\begin{aligned}W &= mgh - 0\\ &= mgh\\ &= \left( {75.0\;{\rm{kg}}} \right) \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {125\;{\rm{m}}} \right)\\ &= 91875\;{\rm{J}}\end{aligned}\)

Hence, the work done against gravity is 91875 J.

Calculation for part (b)

In one revolution, the pedals goes a horizontal distance,

\(\Delta x = 2\pi r\).

Here,\(r\) is the radius of the circle where the pedal rotates. Then,

\(\begin{aligned}r &= \frac{d}{2}\\ &= \frac{{36.0\;{\rm{cm}}}}{2}\\ &= 18.0\;{\rm{cm}}\\ &= 0.18\;{\rm{m}}\end{aligned}\)

Then, the work done during one revolution by the tangential force is calculated as follows:

\(\begin{aligned}{W_{\rm{t}}} &= F\Delta x\\ &= F \times 2\pi r\end{aligned}\)

During one revolution, the cycle goes 5.10 m along the path.Now,

\(\begin{aligned}\frac{{h'}}{x} &= \sin \theta \\h' &= x\sin \theta \end{aligned}\)

Then, the gain in potential energy during one revolution of the pedals is calculated as follows:

\(\begin{aligned}\Delta {\rm{P}}{\rm{.E}} &= mgh'\\ &= mgx\sin \theta \end{aligned}\)

Therefore, from energy conservation, the gain in potential energy is equal to the work done by the tangential force. Then,

\(\begin{aligned}{W_{\rm{t}}} &= \Delta {\rm{P}}{\rm{.E}}\\F \times 2\pi r &= mgx\sin \theta \\F &= \frac{{mgx\sin \theta }}{{2\pi r}}\end{aligned}\)

Now substituting the values in the above equation,

\(\begin{aligned}F &= \frac{{\left( {75.0\;{\rm{kg}}} \right) \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {5.10\;{\rm{m}}} \right) \times \sin {{7.50}^ \circ }}}{{2 \times 3.14 \times \left( {0.18\;{\rm{m}}} \right)}}\\ &= 432.84\;{\rm{N}}\end{aligned}\)

Hence, the tangential force exerted on the pedals is 432.84 N.