Force of Friction: Stopping 1,000kg Car at .8 Coeff.

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SUMMARY

The discussion focuses on calculating the force of friction required to stop a 1,000kg car traveling at 30m/sec with a coefficient of friction of 0.80. The force of friction is determined using the formula Ff = μmg, where μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity (9.81m/sec²). The final calculation confirms that the frictional force acting to stop the car is equal to 7848N. The discussion emphasizes the importance of understanding the relationship between friction, mass, and acceleration in physics.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction and its coefficients
  • Basic knowledge of kinematic equations
  • Ability to perform calculations involving mass and acceleration due to gravity
NEXT STEPS
  • Study the derivation of the equation Ff = μmg in detail
  • Learn about kinematic equations and their applications in real-world scenarios
  • Explore the effects of different coefficients of friction on stopping distances
  • Investigate the role of braking systems in vehicle dynamics
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and the principles of friction in motion.

PhysxMakesMeCry
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A driver of a 1,000kg car is traveling at a speed of 30m/sec when he sees an obstruction in the road. It takes driver .75sec to apply foot to brakes. The car begins to slow down at max rate possible for a coefficient of friction of .80 between the road and tires.

My question is: what is the force of friction acting to stop the car after the brakes are applied?

I would think the equation would be Ff=.8(1000kg) but I thought the force of friction needed to be in Newtons and that doesn't give me a final answer in Newtons.
 
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Nevermind, I forgot I left out 9.81m/sec2
 
Here the frictional force=retarding force. frictional force=f=(coefficient of friction)×(mg)=μmg; if stopping distance=s then v2=u2-2as where a=f/m=μg. as the car finally stops so v=0; put all the values......

Hope you will find this link useful : https://problemsofphysics.wordpress.com/2015/07/06/more-than-50-thought-provoking-problems-on-friction/
 
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