Solving Kinetic Friction: Find Coefficient of Friction from Car Stop

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To find the coefficient of kinetic friction for a car that stops after traveling 40 meters from a speed of 20 m/s, the acceleration can be calculated using the formula a = (v^2 - u^2) / (2s), resulting in an acceleration of 5 m/s². The frictional force is balanced by the mass times acceleration, leading to the equation u_k * m * g = m * a. By simplifying, the coefficient of kinetic friction is determined to be u_k = a/g, which results in a value of 0.5. This calculation effectively demonstrates how to derive the coefficient of friction from the given parameters.
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Homework Statement



A car is moving with a speed of 20m/s when the brakes are applied. The wheels lock (stop spinning). After traveling 40m, the car stops. Determine the coefficient of kinetic friction between the tires and the road.

Homework Equations



Fk=uk x Fn

The Attempt at a Solution



I don't know how to encorporate the given into the equation !
 
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There's another equation you need, one that puts all the given data together.
 
You can apply simple formula...

calculate the acceleration 'a'... by
(square of final velocity)-(square of initial velocity)=2x a x distance

a=20x20/(2x40)
a=5

balance the force
uk x mass x g = mass x a
uk=a/g
uk=5/10=0.5
 
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okay thank you ! i really appreciate it :)
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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