The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.15. 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

First calculate the required acceleration, a, with suitable units of m/s^2.

Vfinal = 64 km/h = 17.78 m/s
Vfinal = sqrt(2 a X)
a = (Vfinal)^2/(2X) = 2.11 m/s^2

Then use Newton's second law, along the direction of motion. If the pushing force is F, the net force in the direction of motion is
Fnet = F + M*g sin6 -M*g*cos6*0.15
= M a
Use that equation to solve for F.

You could also use a conservation of energy method. (Work applied) = F*X
= PE loss + KE gain + (work against friction)

Answer 2

45.326