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
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?
3 answers
87m
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
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
1 answer