Hector drives a pickup truck horizontally at 15.0 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 0.400,what is the minimum stopping distance for the truck so the crate will not slide?
A)28.7 m
B)51.0 m
C)33.6 m
D)44.4 m

Question

Quizplus · Accepted Answer

The minimum stopping distance is given by:$$d = \frac{v^2}{2\mu g}$$where:* v is the initial velocity of the truck* $\mu$ is the coefficient of static friction between the crate and the truck bed* g is the acceleration due to gravityPlugging in the given values, we get:$$d = \frac{(15.0 \ m/s)^2}{2(0.400)(9.81 \ m/s^2)} = 28.7 \ m$$Therefore, the minimum stopping distance for the truck so the crate will not slide is 28.7 m.

Hector Drives a Pickup Truck Horizontally at 15