The coefficient of kinetic friction for a 22 kg bobsled on a track is μ =
0.25
angle θ = 6 degrees
Initial speed u = 0
final speed v = 68 km / h
= 68 * ( 5/ 18 ) m / s
= 18.8888 m / s
Net force F ' = F cos θ - f
where f = frictional force = μ[ mg + F sin θ]
= 0.25 [ 215.6 + 0.1045 F ]
= 53.9 + 0.0261 F
So, F ' = 0.9848 F -53.9 -0.0261 F
= 0.9587 F - 53.9
Accleration a = F ' / m
= ( 0.0435 F - 2.45 )
distance S = 75 m
from the relation v ^ 2 - u ^ 2 = 2aS
356.8 - 0 = 2 ( 0.0435F-2.45) 75
0.0435F-2.45 = 2.378
F = 111 N