1. A 1200 kg car moves along a horizontal road
at speed v0 = 20.1 m/s. The road is wet,
so the static friction coefficient between the
tires and the road is only μs = 0.189 and
the kinetic friction coefficient is even lower,
μk = 0.1323.
The acceleration of gravity is 9.8 m/s^2 .
What is the shortest possible stopping distance for the car under such conditions? Use
g = 9.8 m/s^2 and neglect the reaction time of
the driver.
Answer in units of m.
2. A block accelerates at 3.1 m/s^2 down a plane
inclined at an angle 24.0◦.
Find μk between the block and the inclined plane. The acceleration of gravity is
9.81 m/s^2 .
Help!!!
To find the answers to both questions, we will need to use the equations of motion and the relationship between friction and acceleration. I will guide you through the steps for each question.
Question 1:
Step 1: Determine the stopping distance using the formula s = (v0^2) / (2a), where s is the stopping distance, v0 is the initial velocity, and a is the deceleration (negative acceleration).
Step 2: Calculate the deceleration using the formula a = -μs * g, where μs is the static friction coefficient and g is the acceleration due to gravity.
Step 3: Substitute the value of the deceleration in step 2 into the formula in step 1 to find the stopping distance.
Let's now calculate the stopping distance:
Step 1: s = (v0^2) / (2a)
Given v0 = 20.1 m/s and a = -μs * g, where μs = 0.189 and g = 9.8 m/s^2.
Step 2: Calculate a:
a = -0.189 * 9.8 = -1.8642 m/s^2
Step 3: Substitute the values:
s = (20.1^2) / (2 * (-1.8642))
s = 202.01 / (-3.7284)
s ≈ -54.18 m
The stopping distance for the car, under the given conditions, is approximately 54.18 meters. Since distance cannot be negative, we take the absolute value of the result.
Question 2:
Step 1: Determine the coefficient of kinetic friction using the formula μk = tan(angle), where μk is the coefficient of kinetic friction and angle is the angle of inclination.
Let's now calculate the coefficient of kinetic friction:
Given angle = 24.0° and g = 9.81 m/s^2.
Step 1: μk = tan(angle)
μk = tan(24.0°)
Using a calculator, μk ≈ 0.4456.
So, the coefficient of kinetic friction between the block and the inclined plane is approximately 0.4456.
I hope this helps! If you have any further questions, feel free to ask.
Sure! I can help you with both of these problems. Let's take them step by step.
1. To find the shortest possible stopping distance for the car, we need to determine the maximum value of the friction force that can be applied by the road. Since the car is moving, the kinetic friction coefficient applies.
The friction force (F_friction) can be calculated using the formula:
F_friction = μk * m * g
where μk is the kinetic friction coefficient, m is the mass of the car, and g is the acceleration due to gravity.
Given:
μk = 0.1323
m = 1200 kg
g = 9.8 m/s^2
Substituting these values into the formula, we get:
F_friction = 0.1323 * 1200 kg * 9.8 m/s^2
Simplifying the calculation, we find:
F_friction = 1569.936 N
The friction force acts in the opposite direction to the motion of the car, causing it to decelerate. The deceleration (a) can be calculated using Newton's second law:
F_friction = m * a
Rearranging the equation, we find:
a = F_friction / m
Substituting the known values, we get:
a = 1569.936 N / 1200 kg
Simplifying the calculation, we find:
a = 1.30828 m/s^2
To find the stopping distance (d), we can use the following kinematic equation:
v^2 = v0^2 + 2a * d
Since the car is stopping, the final velocity (v) is zero.
0 = (20.1 m/s)^2 + 2 * 1.30828 m/s^2 * d
Simplifying the equation, we get:
d = - (20.1 m/s)^2 / (2 * 1.30828 m/s^2)
Calculating the result, we find:
d ≈ -154.695 m
However, distance cannot be negative, so the shortest possible stopping distance for the car is approximately 154.695 m.
2. To find the coefficient of kinetic friction (μk) between the block and the inclined plane, we can use the following formula:
μk = tan(θ)
where θ is the angle of the inclined plane.
Given:
θ = 24.0°
Substituting the value into the formula, we find:
μk = tan(24.0°)
Calculating the result, we find:
μk ≈ 0.445041
Therefore, the coefficient of kinetic friction (μk) between the block and the inclined plane is approximately 0.445041.
I hope this helps! Let me know if you have any further questions.