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.
No one has answered this question yet.
(a) Calculate accleration (a)
V^2 = 2 aX. a = V^2/(2X)
Then use
F = ma
for the force required to stop in that distance. It will be in the backwards direction
(b) Add M g sin(4 degrees) to the previous answer. It is the weight component along the direction of motion. It will oincrease the negative stopping force required.
(c) The road applies the stopping force to the tires, when the brakes are applied and wheels do not turn freely.
To determine the force applied to bring the car to rest, we can use the equation:
Force = mass × acceleration
(a) In the case where the road is completely level, the car experiences a deceleration due to the application of the brakes. Since the car comes to rest, its final velocity is 0. The initial velocity is given as 20.0 m/s. The distance covered during deceleration is 140 m. We need to find the acceleration first.
Using the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 × acceleration × distance
0^2 = 20.0 m/s^2 + 2 × acceleration × 140 m
Rearranging the equation to solve for acceleration:
acceleration = (0 - 20.0 m/s^2) / (2 × 140 m)
Now we can calculate the force:
Force = mass × acceleration
Given: mass = 1100 kg,
acceleration = [(0 - 20.0 m/s^2) / (2 × 140 m)],
Substituting the values:
Force = 1100 kg × [(0 - 20.0 m/s^2) / (2 × 140 m)]
The force will be negative since it acts in the opposite direction of motion (opposite to the north).
(b) When the road goes down a hill at a 4-degree angle with respect to the horizontal, we need to consider the force of gravity acting on the car. The force of gravity can be broken down into two components: one along the slope and one perpendicular to the slope. The component along the slope opposes the motion, while the component perpendicular to the slope does not affect the motion.
The force of gravity component along the slope can be calculated as:
Force_gravity = mg × sin(θ)
where m is the mass of the car, g is the acceleration due to gravity (approximated as 9.8 m/s²), and θ is the angle of the slope (in this case, 4 degrees).
The force applied to bring the car to rest will be the sum of the force of gravity component along the slope and the force opposing the motion due to braking.
(c) The force that brings the car to rest is the force of friction between the tires and the road. When the brakes are applied, the friction force acts opposite to the direction of motion, slowing down the car. The force of friction is caused by the interaction between the car's tires and the road surface. The brakes apply a force to the tires, which in turn apply an equal and opposite force on the car due to friction. Therefore, the source of the force that brings the car to rest is the friction between the tires and the road surface.