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**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.