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Ke Min is driving along on a rainy night at 23 m/s when he sees a tire branch lying across the road and slams on the brakes when the branch is 60.0 m in front of him. If the coefficient of kinetic friction between the car's locked tires and the road is 0.41, will the car stop before hitting the branch? The car has a mass of 2400 kg.
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v initial = 23m/s
v final = 0 m/s (a dead stop)
u kinetic = 0.41
m = 2400 kg
u kinetic = F kinetic friction/F normal
F = ma
(v final)^2 = (v initial)^2 + 2ax This is the UAM equation i'll use
Solve for x:
x = [(v final)^2 - (v initial)^2]/(2a)
Solve F = ma for a:
a = F/m
Solve u kinetic = F kinetic friction/Fnormal
F kinetic friction = u kinetic * Fnormal
Since F kinetic friction is the only force acting on the car in the horizontal direction, F kinetic friction = F
Now we can plug this into our other equations
a = (ukinetic * F normal)/m
We still need to find F normal:
F = ma
In this case m = 2400 kg and a = acc. due to grav. = -9.81m/s^2
So we'll just say Fnormal = mg for now
Plugging all of this into the other equation, we get
x = [(v final)^2 - (v initial)^2]/[2*(ukinetic * mg)/m]
That, I can be simplified (by cancelling out the m
x = [(v final)^2 - (v initial)^2]/[2*(ukinetic * g)]
We know all of these values, so (I'm not putting units b/c it looks ugly enough on a computer already)
x = (0^2 - 23^2)/(2 * 0.41 * -9.81) = 66m
Soo, it would hit the branch.
Wow...that took a lot longer than I expected. Anyway, it'll probably look less complicated if you write it out on paper.