The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.20. What force is required to push it down a 6.0° incline and achieve a speed of 64 km/h at the end of 75 m?

Answer 1

Fs = m*g = 22kg * 9.8N/kg = 215.6 N. =
Force of the sled.

Fp = 215.6*sin6 = 22.54 N. = Force parallel to the incline.
Fn = 215.6*cos6 = 214.4 N. = Normal = Force perpendicular to the incline.

Fk = u*Fn = 0.2 * 214.4 = 42.88 N. = Force of kinetic friction.

V = 64km/h = 64000m/3600s = 17.78 m/s.

a = (V^2-Vo^2)/2d
a = (17.78^2-0)/150=2.11 m/s^2

Fap-Fp-Fk = m*a
Fap-22.54-42.88 = 22*2.11
Fap-65.42 = 46.42
Fap = 46.42 + 65.42 = 111.8 N. = Force
applied.