Question

concept simulation 5.2 reviews the concepts that are involved in this problem. a car is safely negotiating an unbanked circular turn at a speed of 17 m/s. the road is dry, and the maximum static frictional force acts on the tires. suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry - road value. if the car is to continue safely around the curve, to what speed must the driver slow the car?

Explanation:

Step1: Identify centripetal - force formula

The centripetal force $F_c$ for a car moving in a circular path is provided by the frictional force $F_f$. On a dry - road, $F_{f1}=F_{c1}$, and $F_{f1}=\mu_s1N$ and $F_{c1}=\frac{mv_1^2}{r}$, where $v_1 = 17$ m/s. On a wet - road, $F_{f2}=F_{c2}$, and $F_{f2}=\mu_s2N$, with $\mu_s2=\frac{1}{3}\mu_s1$. Also, $F_{c2}=\frac{mv_2^2}{r}$.

Step2: Set up the ratio of frictional forces

Since $F_{f1}=\mu_s1N$ and $F_{f2}=\mu_s2N$ and $F_{f2}=\frac{1}{3}F_{f1}$, and $F_{f1}=\frac{mv_1^2}{r}$, $F_{f2}=\frac{mv_2^2}{r}$. We have $\frac{F_{f2}}{F_{f1}}=\frac{\frac{mv_2^2}{r}}{\frac{mv_1^2}{r}}$.

Step3: Substitute the relationship between frictional forces

Since $\frac{F_{f2}}{F_{f1}}=\frac{1}{3}$, then $\frac{\frac{mv_2^2}{r}}{\frac{mv_1^2}{r}}=\frac{1}{3}$. The $m$ and $r$ cancel out, and we get $\frac{v_2^2}{v_1^2}=\frac{1}{3}$.

Step4: Solve for $v_2$

Cross - multiply to get $v_2^2=\frac{v_1^2}{3}$. Take the square root of both sides: $v_2 = v_1\sqrt{\frac{1}{3}}$. Substitute $v_1 = 17$ m/s, so $v_2=17\sqrt{\frac{1}{3}}\approx9.8$ m/s.

Answer:

9.8 m/s