Just getting into friction kinetic and static

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SUMMARY

The discussion centers on calculating the stopping distance of a railroad flatcar loaded with crates, given a coefficient of static friction of 0.30 and an initial speed of 43 km/h. The user initially calculated the acceleration using the formula a = μg, resulting in a value of 2.943 m/s². However, the user overlooked the necessary unit conversion from km/h to m/s for the initial speed, which led to an incorrect calculation of stopping distance. The correct approach requires converting 43 km/h to approximately 11.94 m/s before applying the constant acceleration formula.

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blakester
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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.30 with the floor. If the train is initially moving at a speed of 43 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

I use weird variables I've noticed browsing these forums but here we go
fg = friction with ground
N=normal force

ok so from my free body diagram i got Force of x and y is:
x---- Fs = ma
y----N - mg = ma (which a is 0 since there is not vertical acceleration)

working it down I got N = mg, and using Fg=UN and combining the equations i get

ma=U*mg

divide by m on each side and plug in what i know i get

a=ug
a=.3 * 9.81
a= 2.943

Then using my constant acceleration equations i have Vf^2 - Vo^2=2ad

(0)^2 - (43)^2 = 2(2.943)d
d=314.135

This is the wrong answer apparently though according to the website and I can't for the life of me see what i did wrong. Any help is appreciated thanks! My midterms tomorrow so wish me luck too heh
 
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Hi blakester,

blakester said:
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.30 with the floor. If the train is initially moving at a speed of 43 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

I use weird variables I've noticed browsing these forums but here we go
fg = friction with ground
N=normal force

ok so from my free body diagram i got Force of x and y is:
x---- Fs = ma
y----N - mg = ma (which a is 0 since there is not vertical acceleration)

working it down I got N = mg, and using Fg=UN and combining the equations i get

ma=U*mg

divide by m on each side and plug in what i know i get

a=ug
a=.3 * 9.81
a= 2.943

Then using my constant acceleration equations i have Vf^2 - Vo^2=2ad

(0)^2 - (43)^2 = 2(2.943)d

I think here is the error. Your acceleration is in m/s2, but your speed is in km/hour. You need to do a unit conversion for the initial speed.
 
Oh man such a simple mistake haha! Thanks a lot, I'll be sure to double check my units on my midterm in a few hours
 

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