What is the final speed of the bicyclist rolling downhill?

[jorge0531]

1. A bicyclist on an old bike (combined mass: 89 kg) is rolling down (no pedaling or braking) a hill of height 127 m. Over the course of the 355 meters of downhill road, she encounters a constant friction force of 292 Newton. If her speed at the top of the hill is 8 m/s, what is her speed at the bottom of the hill?



2. 1/2 m(V₁)^2 + mgh₁ = 1/2m(V₂)^2 + mgh₂ + F_frx



3. 1/2(89 kg)(8 m/s)^2 + (89 kg)(9.8 m/s^2)(127 m) = 1/2(89 kg)V₂^2 + (292 N)(355 m)

111125.4 J = 44.5 kg V₂^2 + 103660 J

sqrt(7465.4 J / 44.5 kg) = V₂

V₂ = 12.95 m/s

Somehow I'm doing something wrong because this velocity is wrong. Any help would be greatly appreciated.

 

[Fewmet]

3. 1/2(89 kg)(8 m/s)^2 + (89 kg)(9.8 m/s^2)(127 m) = 1/2(89 kg)V₂^2 + (292 N)(355 m)

111125.4 J = 44.5 kg V₂^2 + 103660 J

I think it is just a transcription error. I am getting 113617.4 J rather than 111125.4 J, and a final velocity of 14.96 m/s.

 

[jorge0531]

you were right :P thank you