this one is kind of hard

Question

A bicyclist can coast down a 7.0 degree hill at a steady 9.5 h^-1 km. If the drag force is proportional to the square fo the speed v, so that Fd = cv^2, calculate

(a) the value fo the constant c and

(b) the average force that must be applied in order to descend the hill at 25 h^-1 km. The mass of the cyclist plus bicycle is 80.0 kg. Ignore other types of friction.

ok for (a) I got 6.7 x 10^-13 m^-1 kg

for (b) I am completly lost the way it's put you would assume there to be zero acceleration so how are you suppose to solve (b)???

Answer 1

wait I think i got it lol
can you tell me if my answer for (a) is right???

Answer 2

ok fixing answer for (a)

Answer 3

ok how is this for (a)

8.2 x 10^-14 m^-1 kg

Answer 4

ok for (a)

13.719 m^-1 kg

Answer 5

ok with correct sig figs (a)

13.7 m^-1 kg

Answer 6

ok i'm stuck on part (b)

Answer 7

part b ...
a velocity of 25 h^-1 km

is it constant velocity were there would be no acceleration if so then there would be no net force...

so i'm lost

Answer 8

ok for (a)

13.7 s^-1 kg

Answer 9

for (b) i got

-.4172 N ;[

Answer 10

ok I know this is right I redid it

(a) 13.7 m^-1 kg

Answer 11

ok for b i got 565 N which was correct with book answer in back

Answer 12

constant speed....so force drag=force gravity

mg*sinTheta=cv^2

c= mgsin7deg/(2.64m/s)^2 So I don't get your answer for a.
check that.

Answer 13

ok for drag force

c = v^-2 Fgx = (2.639 s^-1 m)^-2 95.5 N

Fgx = mg sin theta