The coefficient of sliding friction between rubber tires and wet pavement is 0.50. The
brakes are applied to a 750 kg car traveling 30 m per second, and the car skids to a stop.
What is the magnitude and direction of the force of friction that the road exerts on the car?
What would be the magnitude and direction of the acceleration of the car? Why would it be
constant? How far would the car travel before stopping?
To answer your questions step-by-step:
1. Magnitude and direction of the force of friction:
The magnitude of the force of friction can be calculated using the formula: F_friction = μ * (m * g), where μ is the coefficient of sliding friction, m is the mass of the car, and g is the acceleration due to gravity.
Given:
- Coefficient of sliding friction, μ = 0.50
- Mass of the car, m = 750 kg
- Acceleration due to gravity, g = 9.8 m/s^2
Substituting the given values into the formula:
F_friction = 0.50 * (750 kg * 9.8 m/s^2) = 3675 N (approximately)
Since the car is decelerating, the direction of the force of friction will be opposite to the direction of motion. Therefore, the force of friction will act in the opposite direction of the car's velocity.
2. Magnitude and direction of acceleration:
The magnitude of the acceleration can be calculated using the formula: a = (F_net / m), where a is the acceleration, F_net is the net force acting on the car, and m is the mass of the car.
In this case, the net force is the force of friction (since there are no other forces mentioned).
Substituting the values:
a = (3675 N / 750 kg) = 4.9 m/s^2
The direction of the acceleration will be opposite to the car's initial direction of motion (i.e., in the direction opposite to the velocity vector).
3. Constancy of acceleration:
The acceleration of the car would be constant because the only significant force acting on the car is the force of friction. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Since the net force in this case is constant (force of friction) and the mass of the car remains constant, the acceleration will also be constant.
4. Distance traveled before stopping:
To calculate the distance traveled before stopping, you can use the kinematic equation: v^2 = u^2 + 2as, where v is the final velocity (which will be 0 m/s since the car stops), u is the initial velocity, a is the acceleration, and s is the distance.
Given:
- Initial velocity, u = 30 m/s
- Final velocity, v = 0 m/s
- Acceleration, a = -4.9 m/s^2 (negative sign indicates deceleration)
Substituting the given values into the equation:
0^2 = 30^2 + 2 * (-4.9) * s
0 = 900 - 9.8s
9.8s = 900
s = 900 / 9.8 ≈ 91.84 m
Therefore, the car would travel approximately 91.84 meters before coming to a stop.
To find the magnitude and direction of the force of friction that the road exerts on the car, we can use the formula:
Frictional force = coefficient of friction * Normal force
1. First, we need to calculate the normal force. The normal force is the force exerted by a surface perpendicular to the object in contact with it. In this case, the weight of the car provides the normal force. We can find it using the formula:
Normal force = mass * gravitational acceleration
Given:
mass (m) = 750 kg
gravitational acceleration (g) = 9.8 m/s²
Normal force = 750 kg * 9.8 m/s² = 7350 N
2. Now, we can calculate the frictional force using the coefficient of sliding friction:
Frictional force = 0.50 * 7350 N = 3675 N
The magnitude of the frictional force that the road exerts on the car is 3675 N. The direction of the force of friction is opposite to the direction of motion, which means it acts in the direction opposite to the car's velocity.
To find the magnitude and direction of the acceleration of the car, we can use Newton's second law of motion, which states that:
Force = mass * acceleration
In this case, the frictional force provides the acceleration:
3675 N = 750 kg * acceleration
Solving for acceleration:
acceleration = 3675 N / 750 kg = 4.9 m/s²
The magnitude of the acceleration is 4.9 m/s². The direction of the acceleration is in the direction opposite to the car's initial velocity.
The acceleration would be constant because we assume that the frictional force is the only force acting on the car, and it's assumed to be constant during the car's deceleration.
To find the distance the car travels before stopping, we can use the equation of motion:
v² = u² + 2as
Where:
v = final velocity (0 m/s as the car comes to a stop)
u = initial velocity (30 m/s)
a = acceleration (4.9 m/s²)
s = distance traveled
Rearranging the equation, we get:
s = (v² - u²) / (2a)
Plugging in the values:
s = (0² - 30²) / (2 * -4.9)
s = (-900) / (-9.8) = 91.84 m
Therefore, the car would travel approximately 91.84 meters before coming to a stop.
7350 N
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