Physics Homework - Coefficent of friction.

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SUMMARY

The discussion focuses on calculating the coefficient of friction for two scenarios involving vehicles. In the first scenario, a 2000 kg car sliding to a stop on a snowy road requires the use of kinetic energy loss equated to the work done by friction to find the coefficient. In the second scenario, a 3000 kg car accelerating under a force of 5500 N also necessitates the application of the frictional force formula. The coefficient of friction is defined as the ratio of the frictional force to the normal force, expressed as μk = Ff/Fn.

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Huskies101
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I understand what to do for single step problems, but I get lost when it comes to multi-stepped. Help, please!

1. A 2000 kg car traveling at 20 m/s slides to a stop over a distance of 200 m on a snowy level road. What is this coefficient of friction between car's tires and the road?

2. A 3000 kg car accelerates along a level road at a rate of 1.5 m/s when 5500 N of force is applied. What is its coefficient of friction?

Honestly, I am completely lost, and I do need help. So please, can you help me?
 
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Kinetic energy lost by the car is equal to the work done by the frictional force.
Coefficient of friction = ...?
 
rl.bhat said:
Kinetic energy lost by the car is equal to the work done by the frictional force.
Coefficient of friction = ...?

I know coefficient of friction = us= FnFf. But how do I find Ff? (Frictional force.) Is it the same as the force applied?
 
Frictional force X displacement = Loss of KE
And us = Ff/Fn
 
Kinetic Friction Force=the product of the coefficient of the kinetic friction AND the normal force
"Ffk=μkFn"
 

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