A 1100 kg car is moving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to rest in a distance of 140 m.
(a) If the road is completely level, what is the constant force applied to the car to bring it to rest? (Remember to specify the direction.)
(b) If the road goes down a hill at a 4 degree angle with respect to the horizontal, what is the constant force applied to the car to bring it to rest?
(c) What is the source of the force that brings the car to rest? (In other words, what is exerting the force on the car?) Explain your answer clearly.
To find the answers to the given questions, we need to apply the laws of physics. Let's start by finding the force required to bring the car to rest on a level road.
(a) On a level road, the only force acting on the car is the force of friction between the road and the car's tires. This force opposes the motion of the car and brings it to rest. According to Newton's second law of motion, the force (F) exerted on an object is equal to its mass (m) multiplied by its acceleration (a). In this case, the acceleration is negative since the car is slowing down. The equation can be written as:
F = m * a
Since the car comes to rest, its final velocity (vf) is 0, and its initial velocity (vi) is 20.0 m/s. The displacement (d) is given as 140 m. We can use the kinematic equation:
vf^2 = vi^2 + 2 * a * d
Plugging in the known values:
0^2 = (20.0 m/s)^2 + 2 * a * 140 m
Solving for a:
a = - (20.0 m/s)^2 / (2 * 140 m)
a = - 4 m/s^2
Now we can calculate the force:
F = (1100 kg) * (-4 m/s^2)
F = -4400 N
The negative sign indicates that the force is in the opposite direction of motion, which is north.
Therefore, the constant force applied to the car to bring it to rest on a level road is 4400 N in the opposite direction of motion (north).
(b) If the road goes down a hill at a 4-degree angle with respect to the horizontal, there will be an additional force acting on the car due to gravity. This force will contribute to bringing the car to rest on the downward slope.
The gravitational force (Fg) exerted on the car can be calculated using the equation:
Fg = m * g * sin(theta)
Where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the slope.
Fg = (1100 kg) * (9.8 m/s^2) * sin(4 degrees)
Fg = 427.44 N
On the downward slope, the force of friction (Ff) opposes the motion and contributes to bringing the car to rest. The net force acting on the car is:
Fnet = F - Fg
Fnet = -4400 N - 427.44 N
Fnet = -4827.44 N
The negative sign indicates that the force is uphill, opposite to the direction of motion (north).
Therefore, the constant force applied to the car to bring it to rest on a downward slope with a 4-degree angle is 4827.44 N uphill (opposite to the direction of motion, north).
(c) The force that brings the car to rest is the force of friction between the car's tires and the road. This force is generated due to the roughness of the road surface and the contact between the tires and the road. As the driver applies the brakes, the force of friction counteracts the forward motion of the car, ultimately bringing it to rest.