Question

A cyclists is costing at 12 m/s when she starts down 450 m long slope that is 30

Answer 1

You are posted half the question.

I think the question is asking for the speed of the cyclist at the bottom of the inclined plane.

I assume some extra variables as the problem requires, please edit the values as mentioned in the problem.

Mass of cyclist and her bicycle, m = 70 kg

Drag force due to air resistance acting on the cyclist = 15 N

So, with these variables I am solving the problem.

Inclination of the slope, Θ = arcsin(30/450) = 3.8º

Weight downslope = mgsinΘ = 70kg * 9.8m/s² * sin3.8 = 45.6 N

Friction upslope = 15 N

So, net downslope force = 45.6 - 15 = 30.5 N,

Therefore, net downslope acceleration = 30.5 N / 70kg = 0.43 m/s²

Now, use the following expression -

v² = u² + 2as

= (12m/s)² + 2 * 0.43m/s² * 450m = 531 m²/s²

=> v = 23.04 m/s (Answer)