Question

A car with a mass of 1.2x10^3 kg is skidding to a stop along a horizontal surface. The car decelerates from 32 m/s to rest in 3.94 seconds. Assuming negligible air resistance, determine the coefficient of friction between the car tires and the road surface. Answer: μ =_________ (no units) (rounded to nearest hundredths place)

Answer 1

The coefficient of friction μ is defined as the quotient between the force of friction f and the normal force F_N that an object is subject to:

`\mu=(f)/(F_N)`

Since the friction is responsible for decelerating the car, use Newton's Second Law of Motion to find the magnitude of f:

`f=ma`

The normal force is equal to the weight of the car because the surface is horizontal:

`F_N=mg`

Replacing the expressions for f and F_N, the coefficient of friction becomes:

`\begin{gathered} \mu=(ma)/(mg) \\ \\ \Rightarrow\mu=(a)/(g) \end{gathered}`

The acceleration is the rate of change of the velocity with respect to time. Since the car decelerates from 32m/s to rest in 3.94 seconds, then:

`a=(32(m)/(s))/(3.94s)=8.1218...(m)/(s^2)`

Replace the value of a as well as g=9.81m/s^2 to find the coefficient of friction:

`\begin{gathered} \mu=(a)/(g)=(8.1218...(m)/(s^2))/(9.81(m)/(s^2))=0.8279... \\ \\ \therefore\mu\approx0.83 \end{gathered}`

Therefore, to the nearest hundredths place, the coefficient of friction is 0.83.