Question

question 20
at what maximum speed can a car safely negotiate a turn with a radius of 45 m in icy weather conditions if the coefficient of friction is 0.10? (hint: read the examples in the textbook)
6.6 m/s
0.90 m/s
15 m/s
7.1 m/s

Explanation:

Step1: Identify centripetal - friction force relationship

The centripetal force $F_c$ required for the car to turn is provided by the frictional force $F_f$. So $F_c = F_f$. The centripetal - force formula is $F_c=\frac{mv^{2}}{r}$ and the frictional - force formula is $F_f=\mu N$, where $N = mg$ (on a flat surface, normal force equals weight). So $\frac{mv^{2}}{r}=\mu mg$.

Step2: Solve for the speed $v$

Cancel out the mass $m$ from both sides of the equation $\frac{mv^{2}}{r}=\mu mg$. We get $\frac{v^{2}}{r}=\mu g$. Rearranging for $v$, we have $v = \sqrt{\mu gr}$. Given $\mu=0.10$, $r = 45m$, and $g = 9.8m/s^{2}$.

Step3: Calculate the value of $v$

Substitute the values into the formula: $v=\sqrt{0.10\times9.8\times45}=\sqrt{44.1}\approx6.6m/s$.

Answer:

6.6 m/s