Question

show all you work, including units, on separate paper. follow the problem solving method
this final exam is cumulative and covers material from the entire course.
from the list of choices, select the one best answer
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question 30 of 62
a box of pizza rests in the trunk of your car. you are driving around a corner with a radius of 50 m at 14 m/s. what is the minimum coefficient of
static friction between the pizza box and the trunk of your car required to keep the box from sliding?
0.58
0.40
0.12
0.28

Explanation:

Step1: Identify the physics concept

The problem involves circular motion and static friction. The centripetal force required to keep the box moving in a circle is provided by static friction. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, and the maximum static friction force is $F_f = \mu_s N$, where $N = mg$ (normal force equals weight for horizontal surface).

Step2: Set centripetal force equal to friction force

For the box not to slide, $F_c = F_f$. So, $\frac{mv^2}{r} = \mu_s mg$. The mass $m$ cancels out, giving $\mu_s = \frac{v^2}{rg}$.

Step3: Plug in the values

Given $v = 14\ m/s$, $r = 50\ m$, and $g = 9.8\ m/s^2$. Calculate $\mu_s = \frac{(14)^2}{50\times9.8} = \frac{196}{490} = 0.40$? Wait, no, wait: Wait, $14^2 = 196$, $50\times9.8 = 490$, $196\div490 = 0.4$? But wait, let me recalculate. Wait, $14^2 = 196$, $50\times9.8 = 490$. $196\div490 = 0.4$? Wait, but the options include 0.40. Wait, but let me check again. Wait, maybe I made a mistake. Wait, $v = 14$, $r = 50$, $g = 9.8$. So $\mu_s = v^2/(rg) = 196/(50*9.8) = 196/490 = 0.4$. So the answer is 0.40.

Answer:

0.40 (corresponding to the option with 0.40)