Frictional Forces and Two dimensional constant acceleration problem

AI Thread Summary
The discussion revolves around calculating the stopping distance of a railroad flatcar loaded with crates, considering static friction. The coefficient of static friction is 0.36, and the initial speed of the train is 42 km/h. The participant attempts to use the static frictional force equation and kinematic equations to find the acceleration and stopping distance but arrives at an incorrect result of 250 meters. The need to check units for acceleration and velocity is emphasized to ensure accurate calculations. Proper unit conversion and application of physics principles are crucial for solving this problem correctly.
Zach Lunch
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1. 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 42 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?



2. The Kinematic equations and Static Frictional Force Equation



3. So since the crates aren't moving along the up or down the normal force and gravitational force must be the same. That being said, the static frictional force is the coefficient of static friction times the gravitational force. So I thought I would set that to the force created from the train moving: μ*m*g=m*a. After eliminateing mass I got the acceleration to be 3.528 and then I plugged that into the kinomatic equation: vf^2= (vi^2)+2*a*d and solved for d and got 250. That is incorrect, but I suck at physics and can not do the problems, period.
 
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Zach Lunch said:
I got the acceleration to be 3.528 and then I plugged that into the kinomatic equation: vf^2= (vi^2)+2*a*d and solved for d and got 250. That is incorrect, but I suck at physics and can not do the problems, period.
Check the units. What is the unit of the acceleration and what units you have used for the velocity to get d?
 
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