Hector drives a pickup truck horizontally at 9.00 m/s.He is transporting a crate of delicate lead crystal.If the coefficient of static friction between the crate and the truck bed is 0.740,what is the minimum stopping distance for the truck so the crate will not slide?
A)5.58 m
B)6.66 m
C)81.0 m
D)15.6 m
E)2.79 m

Question

Quizplus · Accepted Answer

To prevent the crate from sliding, the maximum acceleration (a) must not exceed the acceleration due to friction, which can be calculated using the formula $a = \mu g$, where $\mu$ is the coefficient of static friction (0.740) and $g$ is the acceleration due to gravity (9.8 m/s$^2$). Thus, $a = 0.740 \times 9.8 = 7.252$ m/s$^2$. Using the equation $v^2 = u^2 + 2as$ (where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance), and setting $v = 0$ for a complete stop, we solve for $s$: $0 = (9.00)^2 + 2(-7.252)s$. Solving for $s$ gives $s = 5.58$ m, which means the minimum stopping distance to prevent the crate from sliding is 5.58 meters.

Hector Drives a Pickup Truck Horizontally at 9