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
[tex]\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W[/tex]
Where:
m = Mass of the cyclist = 70 kg
W = work done by the drag force = [tex]-F_Dd[/tex]
Where:
d = Distance traveled = 450 m
Therefore,
[tex]\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd[/tex] and
[tex]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}[/tex]
= 658.714 m²/s²
v₂ = 25.665 m/s
Her speed at the bottom of the slope = 25.665 m/s.
