You live at the top of a steep (a slope of ???? degrees above the horizontal) hill and must park your 2200 kg car on the street at night.
a) You unwisely leave your car out of gear one night and your handbrake fails.
Assuming no significant frictional forces are acting on the car, how quickly will it accelerate down the hill?
b) You resolve to always leave your car in gear when parked on a slope. If the rolling frictional force caused by leaving the drive--?train connected to the wheels is 5000 N, at what rate will your car accelerate down the hill if the handbrake fails again?
See prev post
To answer these questions, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
a) In this scenario, assuming no significant frictional forces are acting on the car, the only force accelerating the car down the hill is gravity. The force of gravity can be calculated using the equation:
Force = mass x acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2. The mass of the car is given as 2200 kg.
Force = 2200 kg x 9.8 m/s^2
= 21560 N
Since there are no significant frictional forces, this force will accelerate the car down the hill. To calculate the acceleration, we can use Newton's second law:
Force = mass x acceleration
21560 N = 2200 kg x acceleration
Solving for acceleration:
acceleration = 21560 N / 2200 kg
= 9.8 m/s^2
Therefore, the car will accelerate down the hill at a rate of 9.8 m/s^2.
b) In this scenario, if the rolling frictional force caused by leaving the drive-train connected to the wheels is 5000 N, we need to subtract this force from the force of gravity to calculate the net force accelerating the car down the hill.
Net force = force of gravity - rolling frictional force
In the previous calculation, we found that the force of gravity is 21560 N.
Net force = 21560 N - 5000 N
= 16560 N
Using Newton's second law again, we can find the acceleration:
16560 N = 2200 kg x acceleration
acceleration = 16560 N / 2200 kg
= 7.53 m/s^2
Therefore, if the handbrake fails again and the rolling frictional force is 5000 N, the car will accelerate down the hill at a rate of 7.53 m/s^2.