A cyclist is coasting at 13 m/s when she starts down a 460 m long slope that is 30 m high. The cyclist and her bicycle have a combined mass of70 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?

Express your answer to two significant figures and include the appropriate units.

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Explanation:

Given that

Initial velocity  u=13m/s

Length of slope

L=460m

Height of slope =30m

Mass of cyclist and bike =70kg

Drag force, fictional force=12 N

Final velocity?

Because the system is not isolated, there is some workdone by the drag force.

Therefore,

∆E=W

K.E(f) - K.E(i) + P.E(f) - P.E(i)=W

½mVf² - ½mVi² + mgy(f) - mgy(i)=W

Note, y(f) = 0, the cyclists is already on the floor

½mVf² -½mVi² - mgy(i) = -Fd × d

½×70×Vf² - ½×70×13²-70×9.81×30=-12×450

35Vf²- 5915 - 20601=-5400

35Vf²=-5400+5915+20601

35Vf²=21116

Vf²=21116/35

Vf²=603.314

Vf=√603.314

Vf=24.56m/s

The final velocity is 24.46m/s at the bottom of the track.

could you add the photo?

explanation: