Question

A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom.What is her speed at the bottom of the slope?

Answer 1

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W$

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = $-F_Dd$

Where:

d = Distance traveled = 450 m

Therefore,

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd$ and

$v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh -F_Dd}{ \frac{1}{2}m} = v_1^2 + 2gh -\frac{ 2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}$

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.