The coefficient of sliding friction between a 10kg rubber tire and wet concrete road is 0.43. Determine the time in which a moving object with a velocity of 35mi/h can come to a stop on the road. What distance is covered at this time?
To determine the time it takes for the object to come to a stop on the road, we need to use the equation of motion under constant acceleration. The equation is:
v = u + at
Where:
v = final velocity (0 m/s since the object comes to a stop)
u = initial velocity (35 mi/h = 15.65 m/s, after converting to m/s)
a = acceleration
t = time
Now, we need to find the acceleration. The acceleration can be calculated using the equation:
F_friction = μ * N
Where:
F_friction = force due to friction
μ = coefficient of sliding friction (given as 0.43)
N = normal force
The normal force can be calculated using the equation:
N = mg
Where:
m = mass of the object (10 kg)
g = acceleration due to gravity (9.8 m/s^2)
Now, let's calculate the acceleration:
N = mg
N = 10 kg * 9.8 m/s^2
N = 98 N
F_friction = μ * N
F_friction = 0.43 * 98 N
F_friction = 42.14 N
Since the friction force is in the opposite direction of motion, it will provide a negative acceleration, therefore:
a = -F_friction/m
a = -42.14 N / 10 kg
a = -4.214 m/s^2
Now, we can substitute the values in the first equation and solve for time (t):
0 = 15.65 m/s + (-4.214 m/s^2) * t
Rearranging the equation gives:
4.214 m/s^2 * t = 15.65 m/s
t = 15.65 m/s / 4.214 m/s^2
t ≈ 3.71 seconds
So, it will take approximately 3.71 seconds for the object to come to a stop on the road.
To determine the distance covered at this time, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
Plugging in the values:
s = 15.65 m/s * 3.71 s + (1/2) * (-4.214 m/s^2) * (3.71 s)^2
s ≈ 58.10 m
Therefore, the object will cover approximately 58.10 meters before coming to a stop on the road.