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?

m

F=ma

mu*mg=ma
solve for acceleration.

Now, using

vf^2=vi^2+2ad solve for distance d.

To find how short a distance the train can stop without causing the crates to slide over the floor, we need to determine the maximum deceleration the train can experience without exceeding the static friction between the crates and the floor.

Step 1: Convert the initial speed of the train to meters per second.
Given:
Initial speed of the train = 47 km/h

Conversion factor: 1 km/h = 1000 m / 3600 s

Calculation:
Initial speed of the train = 47 km/h * (1000 m / 3600 s) = 13.0556 m/s (approximately)

Step 2: Calculate the maximum deceleration the train can experience without the crates sliding.
Given:
Coefficient of static friction (μ) = 0.20

Formula:
Maximum deceleration (a) = μ * g

where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Calculation:
Maximum deceleration (a) = 0.20 * 9.8 m/s^2 = 1.96 m/s^2

Step 3: Calculate the stopping distance of the train.
Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the stopping distance.

Given:
Initial velocity (u) = 13.0556 m/s
Final velocity (v) = 0 m/s (since the train is stopping)
Acceleration (a) = -1.96 m/s^2 (negative sign because the train is decelerating)
Stopping distance (s) = ?

Rearranging the equation:
s = (v^2 - u^2) / (2a)

Calculation:
s = (0^2 - 13.0556^2) / (2 * -1.96)
s = (-170.1389) / (-3.92)
s ≈ 43.4 m (approximately)

Therefore, the train can be stopped in approximately 43.4 meters without causing the crates to slide over the floor.