Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
You are driving a 2590.0 kg car at a constant speed of 14.0 m/s along an icy, but straight, level road. As you approach an inte...Asked by James
you are driving a 2520.0-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 25.8 m. What is the coefficient of kinetic friction between your tires and the wet road?
Answers
Answered by
Henry
Wc = ma = 2520kg * 9.8N/kg = 24,696N =
Weight of car.
Fc = (24,696N.,0 deg.).
Fv = 24,696cos(0) = 24,696N. = Force perpendicular to road.
a = (Vf^2 - Vo^2) / 2d,
a = (0 - (14)^2) / 51.6 = -3.8m/s^2.
Fn = ma = 2520 * (-3.8) = -9572N.
Ff = u*Fv = -(-9572),
24,696u = 9572,
u = 0.388. = Coefficient of friction.
Weight of car.
Fc = (24,696N.,0 deg.).
Fv = 24,696cos(0) = 24,696N. = Force perpendicular to road.
a = (Vf^2 - Vo^2) / 2d,
a = (0 - (14)^2) / 51.6 = -3.8m/s^2.
Fn = ma = 2520 * (-3.8) = -9572N.
Ff = u*Fv = -(-9572),
24,696u = 9572,
u = 0.388. = Coefficient of friction.
Answered by
khaled
where did u get 51.6?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.