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

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

Step-by-step explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

`(1)/(2)mv_2^2 -(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,

`(1)/(2)mv_2^2 -(1)/(2)mv_1^2 - mgh = -F_Dd` and

`v_2^2 =( (1)/(2)mv_1^2 + mgh -F_Dd)/( (1)/(2)m) = v_1^2 + 2gh -( 2F_Dd)/( m) = 15^2 + 2* 9.8* 30 - (2* 12* 450)/(70)`

= 658.714 m²/s²

v₂ = 25.665 m/s

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