Traveling at a speed of 18.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.580. What is the speed of the automobile after 1.44 s have elapsed? Ignore the effects of air resistance.
vf=vi+at
where
a=force/mass=-mu*weight/(weight/g)
= -mu*g
To solve this problem, we can use the equation of motion:
Vf = Vi + at
Where:
Vf is the final velocity (speed) of the automobile
Vi is the initial velocity (speed) of the automobile
a is the acceleration
t is the time elapsed
First, we need to find the acceleration using the equation of motion that relates acceleration and friction:
F = μ * N
Where:
F is the frictional force
μ is the coefficient of kinetic friction
N is the normal force, which is the weight of the automobile in this case
Since the problem states that the automobile suddenly locks the wheels, the frictional force is equal to the maximum static frictional force. Thus, we can write:
F = μs * N
Now, we can substitute the equation F = μs * N into Newton's second law of motion:
F = m * a
Where:
m is the mass of the automobile
a is the acceleration
Equating the above two equations, we get:
μs * N = m * a
Since the normal force (N) is equal to the weight (mg), we can rewrite the equation as:
μs * mg = m * a
Simplifying the equation, we find:
a = μs * g
Now, substituting the acceleration value (a) into the equation of motion:
Vf = Vi + at
Vf = Vi + (μs * g) * t
Finally, we can calculate the final velocity (Vf) by substituting the given values into the equation:
Vf = 18.0 m/s + (0.580 * 9.8 m/s^2) * 1.44 s
Calculating the above expression, we find the final velocity (Vf).