- #1
preet
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I need to find the coeffecient of kinetic friction for the following problem... all the stuff after the dashes is what I've done... The answer is approximately half the answer I keep getting, so I think I'm missing something...
The driver of a 1200kg truck traveling 45km/h [E] on a slippery road applies the brakes, skidding to a stop in 35m. Determine the coeffecient between the road and the car tires.
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45km/h = 12.5m/s
find the time it took to stop to find acceleration:
v=d/t 12.5 = 35/t t=2.56s
find acceleration:
a=v/t a=12.5/2.56
a=4.88 m/s^2
find force:
a=f/m 4.88= f / 1200 f=5856N[E]
find normal force:
mg = 1200 (9.81) = 11772N
find friction coeff.
Force of friction = coeff * Normal Force
5856N = x * 11772
x = .49
The answer in the book is .23... I don't understand how to find friction force. I substituted the force of the truck for the friction because I remember seeing that somewhere with a similar question. I'd appreciate any help.
The driver of a 1200kg truck traveling 45km/h [E] on a slippery road applies the brakes, skidding to a stop in 35m. Determine the coeffecient between the road and the car tires.
------------------------
45km/h = 12.5m/s
find the time it took to stop to find acceleration:
v=d/t 12.5 = 35/t t=2.56s
find acceleration:
a=v/t a=12.5/2.56
a=4.88 m/s^2
find force:
a=f/m 4.88= f / 1200 f=5856N[E]
find normal force:
mg = 1200 (9.81) = 11772N
find friction coeff.
Force of friction = coeff * Normal Force
5856N = x * 11772
x = .49
The answer in the book is .23... I don't understand how to find friction force. I substituted the force of the truck for the friction because I remember seeing that somewhere with a similar question. I'd appreciate any help.