Frictional Forces and Two dimensional constant acceleration problem

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SUMMARY

The discussion centers on calculating the stopping distance of a railroad flatcar loaded with crates, given a coefficient of static friction of 0.36 and an initial speed of 42 km/h. The participant attempted to derive the acceleration using the static frictional force equation, resulting in an acceleration of 3.528 m/s². However, the calculation for stopping distance using the kinematic equation vf² = vi² + 2ad was incorrectly solved, leading to an erroneous distance of 250 meters. The discussion emphasizes the importance of unit consistency in physics calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf² = vi² + 2ad
  • Knowledge of static friction and its calculation using the coefficient of static friction
  • Familiarity with Newton's second law of motion
  • Basic unit conversion skills, particularly between km/h and m/s
NEXT STEPS
  • Review the derivation and application of the static frictional force equation
  • Practice solving kinematic problems involving constant acceleration
  • Learn about unit conversions between different speed measurements, such as km/h to m/s
  • Explore real-world applications of friction in physics, particularly in transportation scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and kinematics, as well as educators seeking to clarify concepts related to friction and motion.

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?
 
Thanks
 

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