by rum2563
Tags: coefficient, friction
 P: 103 The question: The driver of a 1.2X10^3 kg car travelling 45 km/h [W] on a slippery road applies the brakes, skidding to a stop in 35 m. Determine the coefficient of friction between the road and the car tires. Equations: - Ff = uFn - d = 0.5 (Vi + Vf) * time - Fn = mg - F = ma My attempt: To find normal force: Fn = (1.2 * 10^3) (9.81) = 11772 N To find acceleration: d = 0.5 (Vi + Vf) * time 35 = 0.5 (45 + 0) * time 35 = 0.5 (45) * time 35 = 22.5 * time 35/22.5 = time time = 1.6 seconds acceleration = (Vf - Vi) / time = (0 - 45) / 1.6 = -28.12 m/s^2 To find force using Newtons' law: F= (1.2*10^3)(.28.12) = -33744 N u (coefficient of friction) = -33744 / 11772 = -2.87 Is the answer correct? Basically we have to find the coefficient of friction, but I am confused as to how to find it out. The main problem is that acceleration is negative, can anyone please help. This is due tomorrow for me. Thanks.
 P: 35 watch out for 45km/h...
P: 103
 Quote by 21385 watch out for 45km/h...
Could you please elaborate a bit more? Thanks.

P: 35

divide by 3.6 to get meters per second
P: 103
 Quote by 21385 divide by 3.6 to get meters per second
But I am confused. Shouldn't I divide by 1.6 because that is the time? Or is 3.6 another number? Thanks for your help, please explain.
Mentor
P: 40,707
 Quote by rum2563 To find acceleration: d = 0.5 (Vi + Vf) * time 35 = 0.5 (45 + 0) * time 35 = 0.5 (45) * time 35 = 22.5 * time 35/22.5 = time time = 1.6 seconds acceleration = (Vf - Vi) / time = (0 - 45) / 1.6 = -28.12 m/s^2
Careful with units! Convert the car speed to m/s.
Nothing wrong with a negative acceleration--the car is slowing down. The sign merely specifies direction--when computing the coefficient, consider magnitudes.
 P: 35 The standard units for velocity is m/s. The velocity used in the question is in km/h. The other units used in the question is in meters and naturally, you would want meters per second. You have to convert the velocity to get a correct answer. To convert 45 km/h (1000m/1km)(1 h/3600 seconds) ===> 45 km/h divided by 3.6 would get you 12.5 m/s, which is what you want.
 P: 103 So when I convert 45 km/h to m/s, I get 12.5 m/s So do I use 12.5m/s to get the acceleration just like I did in my question?
 P: 35 yep, that should give you the correct answer
 P: 103 One more question, when I do all the calculations, I get a negative acceleration. So therefore, I would get a negative value for coefficient of friction. Is that OK, or would it not make sense because friction cannot be negative. This is what I am doing: u = -9378 / 11772 = -0.8 Please help, thanks.
 Mentor P: 40,707 Friction, a force, can certainly be "negative". Remember: sign just refers to direction. But the coefficient of friction cannot be negative: Ff = uFn describes the magnitude of the friction force, not the direction or sign.
P: 1,345