Posted by **johnny** on Thursday, December 11, 2008 at 9:54pm.

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.20 with the floor. If the train is initially moving at a speed of 47 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

- physics -
**drwls**, Friday, December 12, 2008 at 6:42am
47 km/h = 13.056 m/s

Compute the acceleration(a)when the stopping distance is X meters:

V^2 = 170.44 m^2/s^2 = 2 a X

a = V^2/(2X) = 85.22/X

M g * us = M a when a is as high as possible without slipping

M's cancel. You know the static fricion coefficient us. Substitute for a and solve for X

- physics -
**billy**, Thursday, March 7, 2013 at 5:02pm
87m

- physics -
**lilly**, Tuesday, October 20, 2015 at 4:24am
coefficient of static friction = 0.20

fs = static frictional force

mass = m

g = 9.8m/s^2

as we know that fs = fnormal*coefficient of static friction

so, fs= mg*0.20

as fnormal= mg = 9.8m N

now, acc to newtons law

fnet = fs

ma = 9.8*0.20*m

m's get cancel

a= 1.96m/s^2

applying the formula

v^2= u^2 + 2ax

now substitute the value in this equation

you get, x= 43.1m

don't forget to change speed into m/s

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