Car traveling on horizontal road with force of static friction

AI Thread Summary
A car traveling at 50 mph on a horizontal road needs to calculate stopping distances under different friction conditions. For a rainy day with a static friction coefficient of 0.100, the stopping distance can be determined using the equation of motion that relates distance, initial velocity, and acceleration. The static friction force is equal to the coefficient multiplied by the normal force, which leads to the acceleration needed for stopping. When the surface is dry with a coefficient of 0.600, the stopping distance will be significantly shorter due to increased friction. Understanding the relationship between net force and static friction is crucial for solving these types of problems.
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Homework Statement


A car is traveling at 50 mi / hr on a horizontal highway

a. If the coefficient of static friction between the road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

b. What is the stopping distance when the surface is dry and μs = 0.600?

Homework Equations


∑ f = ma

fs = (μs)(normal force)

Xf = Xi + (Vi)(t) + (1/2)at2


The Attempt at a Solution


Do not know where to start, please help.

Part a:

∑ Fx = ma - fs = 0

fs = (μs)(normal force)

∑ Fy = (normal force) = mg

ma = fs therefore ma = (μs)(normal force)

a = (μs)(normal force) / mass

Now: vf = 0

vi = 73.3 ft / second

Xf = Xi + Vi t + (1/2)at2

0 = 50 mi / hr + [ (0.100)(normal force) / mass ][t2]
 
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Color_of_Cyan said:

Homework Statement


A car is traveling at 50 mi / hr on a horizontal highway

a. If the coefficient of static friction between the road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

b. What is the stopping distance when the surface is dry and μs = 0.600?

Homework Equations


∑ f = ma

fs = (μs)(normal force)

Xf = Xi + (Vi)(t) + (1/2)at2


The Attempt at a Solution


Do not know where to start, please help.
Actually, you have started off pretty well.
Part a:

∑ Fx = ma - fs = 0
Yes, but try to avoid the use of pseudo inertial forces, where practical, when using Newton's 2nd law. Rather use F_net = ma, where F_net in this case is fs, in the direction of the acceleration. It's the same result as F_net = ma - fs = 0, but it avoids confusion down the road.
fs = (μs)(normal force)

∑ Fy = (normal force) = mg

ma = fs therefore ma = (μs)(normal force)

a = (μs)(normal force) / mass

Now: vf = 0

vi = 73.3 ft / second

Xf = Xi + Vi t + (1/2)at2

0 = 50 mi / hr + [ (0.100)(normal force) / mass ][t2]
Rather than use this kinematic equation that relates distance, acceleration, initial velocity, and time, use the kinematic equation of motion that relates distance, initial and final velocity, and acceleration. That avoids solving for the time.
 
That helped a lot, thanks!

I didn't know you were supposed to set netforce equal to force of static friction.
 
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