Question

quiz instructions
you may complete this review up to 3 times. you can earn up to 10 points of extra credit on the unit 4 quiz.
round all numerical answers to the fourth digit behind the decimal place. enter all numerical answers in standard
floating notation (normal decimals), not scientific notation.
question 6
a car turns on a flat, curved stretch of highway that has a radius of 227 meters. the car can
ravel at a maximum velocity of 32 m/s before it starts to slip off of the road. what must be
the coefficient of friction between the cars tires and the road?

Explanation:

Step1: Equate centripetal force and frictional force

The centripetal force \(F_c=\frac{mv^{2}}{r}\) and the frictional force \(F_f = \mu N\). On a flat road, \(N = mg\). So, \(\frac{mv^{2}}{r}=\mu mg\).

Step2: Solve for the coefficient of friction \(\mu\)

Cancel out \(m\) from both sides of the equation \(\frac{mv^{2}}{r}=\mu mg\). We get \(\mu=\frac{v^{2}}{rg}\). Given \(v = 32\space m/s\), \(r=227\space m\), and \(g = 9.8\space m/s^{2}\). Substitute the values: \(\mu=\frac{32^{2}}{227\times9.8}\).

Step3: Calculate the value

First, calculate \(32^{2}=1024\), then \(227\times9.8 = 2224.6\). So, \(\mu=\frac{1024}{2224.6}\approx0.459\).

Answer:

\(0.459\)