How short a distance to come to a stop without sliding?

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SUMMARY

The discussion focuses on calculating the stopping distance of a railroad car containing crates, given a coefficient of static friction of 0.5 and an initial velocity of 22 m/s. The net force is zero to prevent sliding, leading to an acceleration of 4.9 m/s² derived from the static friction multiplied by gravity. Participants emphasize using kinematic equations to relate initial velocity, acceleration, and distance to determine how short the stopping distance can be without the crates sliding.

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A railroad car has crates in it with a coefficient of .5 w/ the car's floor. The train is moving at 22m/s, how short a distance can it be stopped w/o letting the crates slide?


Fnet = 0 because you don't want sliding.
a = coefficient of static friction x gravity



When I solved for a, I got 4.9m/s2. I know if I multiply this by the mass of the crates, it should be equal to the static friction force.

I think I have to take the derivative of the velocity to get acceleration, and the antiderivative of velocity to get distance, but how do I do this with no other variables?
 
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I've not checked the working for the earlier part of the question.

I think I have to take the derivative of the velocity to get acceleration, and the antiderivative of velocity to get distance, but how do I do this with no other variables?

Draw sketches of the train at the start and the end of the problem, write on all the values you know, and see what equations you might be able to use. (Think kinematics.)
 
You have the initial velocity and the acceleration. You can find distance traveled from that information.
 

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