the floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.36 with the floor. if the train is initially moving at a speed of 45 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

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21-10-2022
Physics

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the floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.36 with the floor. if the train is initially moving at a speed of 45 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Answer :

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tutorconsortium115 tutorconsortium115 · Answer 1 · 2022-10-28T01:18:59-04:00

24.8 m is the shortest braking distance.

On the horizontal axis, we dread the force of friction (fr) that prevents the train from moving and accelerating.

Y axis

N- W = 0

N = W = mg

explanation?

In order to fix this issue, we must apply Newton's second law to the boxes, with the norm vertically up and the weight vertically down.

Write the equation for each axis. On the horizontal axis, we dread the force of friction (fr) that prevents the train from moving and accelerating.

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.36

a = - 3.14 m/s²

Utilizing the kinematics equations, we determine the distance.

(Vf2 - Vo2) / 2 a = Vf2 = Vo2 + 2 an

The speed of the train is zero (Vf = 0) when it stops.

Vo=45 km/h (1000 m/km) (1 h 3600 s)= 12.5m/s

x = ( 0 - 12.5²) / 2 (-3.14)

x= 24.88.8 m

The shortest braking distance is 24.8 m

To learn more about the distance problem refer to: https://brainly.com/question/13371241

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