A bicyclist can coast down a 5° hill at a constant 6.0 km/h. Assume the force of friction (air resistance) is proportional to the speed v so that Ffr = cv.

(a) Calculate the value of the constant c.

(b) What is the average force that must be applied in order to descend the hill at 20 km/h? The mass of the cyclist plus bicycle is 77 kg.

To solve this problem, we will first need to calculate the constant c in part (a), and then use that value to find the average force in part (b).

(a) To calculate the value of the constant c, we can use the information given: the bicycle is coasting down a 5° hill at a constant speed of 6.0 km/h.

We know that the force of friction (Ffr) is proportional to the speed (v), so we can express this relationship as Ffr = cv.

At 6.0 km/h, we need to convert the speed to m/s since the SI unit for force is Newtons. 1 km/h is equivalent to 1000 m/3600 s, which simplifies to 0.2778 m/s.

Now, let's consider the gravitational force acting on the bicycle. The force of gravity can be calculated using the equation Fg = mg, where m is the mass of the cyclist plus bicycle and g is the acceleration due to gravity. In this case, m = 77 kg, and g is approximately 9.8 m/s².

Since the bicycle is coasting down a 5° hill, we can decompose the gravitational force into two components: the component of the force parallel to the hill (Fp), which opposes the force of friction, and the component of the force perpendicular to the hill (Fn), which sustains the weight of the cyclist and bicycle.

The parallel component of the gravitational force is Fp = mg sin(θ), where θ is the angle of inclination of the hill. In this case, θ = 5°.

The force of friction (Ffr) acting on the bicycle when coasting down the hill must balance the parallel component of the gravitational force (Fp), so we can set them equal to each other:

cv = mg sin(θ)

Plugging in the known values, we get:

c * 0.2778 = 77 * 9.8 * sin(5°)

Now, we can solve for c:

c = (77 * 9.8 * sin(5°)) / 0.2778

Using a calculator, we find that c ≈ 111.076.

Therefore, the value of the constant c is approximately 111.076.

(b) To find the average force that must be applied to descend the hill at a speed of 20 km/h, we can use the same equation as before: Ffr = cv.

At 20 km/h, the speed is 20 * (1000/3600) m/s, which simplifies to 5.56 m/s.

Now, we can calculate the force of friction at this speed:

Ffr = 111.076 * 5.56

Using a calculator, we find that Ffr ≈ 617.07 N.

Since the force of friction (Ffr) is equal to the average force applied to descend the hill, the average force required to maintain a speed of 20 km/h is approximately 617.07 N.

Therefore, the average force that must be applied to descend the hill at 20 km/h is approximately 617.07 Newtons.