Just getting into friction kinetic and static

AI Thread Summary
The discussion revolves around calculating the stopping distance of a railroad flatcar loaded with crates, where the coefficient of static friction is 0.30. The user initially calculates the acceleration due to friction as 2.943 m/s², using a free body diagram and equations of motion. However, the error identified is that the initial speed of 43 km/h needs to be converted to meters per second for accurate calculations. This oversight affects the final stopping distance result, highlighting the importance of unit consistency in physics problems. The user expresses gratitude for the clarification and plans to double-check units for their upcoming midterm.
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 alot, I'll be sure to double check my units on my midterm in a few hours
 
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