Kinetic Friction Problem (Truck Braking)

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Spartan Erik
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Homework Statement



"A bread truck is traveling at 18 m/s on a horizontal road. The brakes are applied and the truck skids to a stop in 3.6s. The coefficient of kinetic friction between the tires and the road is:"

Homework Equations



Kinetic Friction = coefficient of kinetic friction x normal force

The Attempt at a Solution



I'm not sure how you can manipulate velocity and time to get kinetic friction..
 
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Hint: What's the truck's acceleration?
 
Well if brakes are being applied, the truck is decelerating until velocity is 0
 
Well acceleration is in units of m/s^2, so 18 m/s / 3.6s = 5 m/s^2 if I'm not mistaken
 
Good! Now make use of it. Hint: What does Newton's 2nd law tell you?
 
Well F = MA, but I can't solve for force or mass.. unless of course there's something big that I'm missing..
 
Make use of what you had in your first post. The "F" is the kinetic friction force.
 
Alright so far I've established that the acceleration is 5 m/s^2, the velocity was 18 m/s, and that the time to stop was 3.6s.

Since Kinetic Friction Force = coefficient of kinetic friction x normal force

I can break the normal force into mass and acceleration, giving the equation
Kinetic Friction Force = coefficient of kinetic friction x mass x acceleration (5 m/s^2)

But I still don't see what unknowns I can solve for..
 
The normal force does not equal "mass X acceleration". To find the normal force add up the vertical components of all forces and set it equal to zero.
 
Oh man I feel like beating myself over the head with a stick right about now.. quite a eureka moment.

Alright so: kinetic frictional force = coefficient of k x normal force
coefficient of k = kinetic frictional force / normal force

Based off that, mass is irrelevant, so it is ultimately acceleration / gravity

So the coefficient is 0.51.. man.. thanks for spurring the ole noggin!