I am pretty sure the answer might be A. Can someone help by showing the work needed to solve the problem. It would be greatly appreciated

1. The problem statement, all variables and given/known data

A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide?

2. Relevant equations

A. 0.400
B. 0.365
C. 0.892
D. 0.656

3. The attempt at a solution

I'm not really sure how to start

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Hi redhawks,

 Quote by redhawks I am pretty sure the answer might be A. Can someone help by showing the work needed to solve the problem. It would be greatly appreciated 1. The problem statement, all variables and given/known data A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide? 2. Relevant equations A. 0.400 B. 0.365 C. 0.892 D. 0.656 3. The attempt at a solution I'm not really sure how to start

Draw a force diagram for the crate by itself, showing all forces that act on it directly and its acceleration. Then use newton's laws to get two equations, one for the horizontal force components and one for the vertical force components.

 Thanks, but i know to do the FBD. I think the distance part throws me off. If someone can go through the steps, it would really help me out.

i arrived at solution a also.... although i cant give you a written out explanation, i can try to point you in the write direction...

a good first step is to draw out that FBD... for y components you should see you have the weight of the box (downward) and the normal force on the bottom of the box (upward). the x components should show the force of the box due to the acceleration (to the right), and the force friction countering the force of the box (to the left)

you need to figure out the force of the box due to accelartion
use
F=ma, and i used Vf^2-Vi^2=2ad to determine the acceleration a (here i will give you a little tip, i assumed the truck came to a stop after the 28.7m, so that Vf=0) This part of the problem is dynamic when looking at the problem in reference to the truck in motion on the road....

if you then focus your attention to just the box and the bed of the truck, there is no relative motion between the two and you can assume static equations...

here is used sum of x components F=0, i also used (Force of friction)=(normal force)(static coefficient)

let me know how it goes from here, if you need more help i will be glad to, well...., help hahahah

 Thanks a lot. I GOT IT!!!