Asked by brian
A railroad flatcar is loaded with crates having a coefficient of static friction of 0.394 with the floor. If the train is moving at 57.1 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide?
Answers
Answered by
Henry
Vo = 57.1km/h = 57100m/3600s = 15.9 m/s.
Fc = m*g = Force of the crate.
Fs = u*mg = 0.394mg
Fap-Fs = m*a
Fap-0.394mg = m*0 = 0
Fap = 0.394mg = Force applied.
a=Fap/m = 0.394mg/m=0.394g=0.394*(-9.8)
= -3.86 m/s^2.
V^2 = Vo + 2a*d
d = (V^2-Vo^2)/2a = 0-(15.9^2)/-7.72 =
32.7 m.
Fc = m*g = Force of the crate.
Fs = u*mg = 0.394mg
Fap-Fs = m*a
Fap-0.394mg = m*0 = 0
Fap = 0.394mg = Force applied.
a=Fap/m = 0.394mg/m=0.394g=0.394*(-9.8)
= -3.86 m/s^2.
V^2 = Vo + 2a*d
d = (V^2-Vo^2)/2a = 0-(15.9^2)/-7.72 =
32.7 m.
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