KE=workbraking
1/2 m v^2=mu*mg*distance
solve for mu.
You are driving a 2530.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 23.6 m. What is the coefficient of kinetic friction between your tires and the wet road?
2 answers
F = m a = rate of change of momentum
first work on a
average speed during braking = (14+0)/2
= 7 m/s
so
time to stop = 23.6 m / 7 m/s = 3.37 seconds
change of momentum = -2530*14 =-35420
so
F = -35420 / 3.37 = -10510 Newtons
mu m g = 10510
mu = 10510 / (2530*9.81)
= .423
good tires for wet road.
first work on a
average speed during braking = (14+0)/2
= 7 m/s
so
time to stop = 23.6 m / 7 m/s = 3.37 seconds
change of momentum = -2530*14 =-35420
so
F = -35420 / 3.37 = -10510 Newtons
mu m g = 10510
mu = 10510 / (2530*9.81)
= .423
good tires for wet road.
1 answer