### In a pickup game of dorm shuffleboard, students c…

02:11
AH
Carnegie Mellon University
Problem 1

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.25 with the floor. If the train
is initially moving at a speed of 48 $\mathrm{km} / \mathrm{h}$ , in how short a distance
can the train be stopped at constant acceleration without causing
the crates to slide over the floor?

36$m$

## Discussion

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## Video Transcript

so here the greatest deceleration is provided by the maximum frictional force in this case, the normal forces equaling MG. And this is simply from the from the figure and the Newton. Using Newton's second law, we can say that the acceleration would be equaling the maximum static frictional force divided by and the Mass. And this would be equaling to the mat to the er coefficient of static friction multiplied by the normal force or, in this case, MG divided by EMS, the EMS cancel out. And this is giving us this coefficient of static friction multiplied by the acceleration due to gravity. And we can say that the shortest distance to stop would be that magnitude of this of the displacement in the ex direction. This would be equaling the velocity squared, divided by two times the acceleration. And in this case, it's me. The velocity squared, divided by two times the coefficient of static friction times G. And in this case we have a velocity in kilometers per hour. So we'll say 48 kilometers per hour, multiplied by 1000 meters for every one kilometer multiplied by one hour for every 3600 seconds. We're going to square this term. This will be divided by two multiplied by the coefficient of static friction of 20.25 multiplied by 9.8 meters per second squared. And we find that the magnitude of the displacement in the ex direction is going to be equaling two, approximately 36 meters. That is the end of the solution. Thank you for watching.