Question

question 19
at what maximum speed can a car safely negotiate a turn with a radius of 45 m in dry weather conditions if the coefficient of friction is 0.95? (hint: read the examples in the textbook)
4.7 m/s
9.8 m/s
20 m/s
12 m/s

Explanation:

Step1: Identify centripetal - friction force relationship

The centripetal force $F_c$ for the car to make the 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 $F_f=\mu N$. On a level road, $N = mg$. 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 $\frac{v^{2}}{r}=\mu g$. Then $v^{2}=\mu gr$. Given $\mu = 0.95$, $r = 45m$ and $g = 9.8m/s^{2}$, we substitute these values: $v^{2}=0.95\times9.8\times45$. Calculate $0.95\times9.8\times45=0.95\times441 = 418.95$. Then $v=\sqrt{418.95}\approx20.47m/s\approx20m/s$.

Answer:

C. 20 m/s