A 3.5 × 103

kg car accelerates from rest at
the top of a driveway that is sloped at an angle of 16.9◦ with the horizontal. An average frictional force of 4.1×103 N impedes the car’s
motion so that the car’s speed at the bottom of the driveway is 4.2 m/s.
The acceleration of gravity is 9.81 m/s^2.
What is the length of the driveway?
Answer in units of m

See previous post: 10-17-13,5:26 PM.

To find the length of the driveway, we can use the equations of motion. The key equation we will use is:

v^2 = u^2 + 2as

Where:
- v is the final velocity of the car (4.2 m/s)
- u is the initial velocity of the car (0 since it starts from rest)
- a is the acceleration of the car
- s is the distance traveled

First, let's calculate the acceleration of the car. The net force acting on the car is given by:

F_net = m * a

In this case, the net force is the difference between the force component down the slope and the frictional force:

F_net = mg * sin(θ) - F_friction

Where:
- m is the mass of the car (3.5 × 10^3 kg)
- g is the acceleration due to gravity (9.81 m/s^2)
- θ is the angle of the slope (16.9°)
- F_friction is the frictional force (4.1 × 10^3 N)

Let's plug in the values and calculate the net force:

F_net = (3.5 × 10^3 kg) * (9.81 m/s^2) * sin(16.9°) - (4.1 × 10^3 N)
= 4929.63 N - 4100 N
= 829.63 N

Now, we can calculate the acceleration using Newton's second law:

F_net = m * a

a = F_net / m
a = (829.63 N) / (3.5 × 10^3 kg)
a ≈ 0.237 m/s^2

Now, let's substitute the values of u, v, and a into the equation of motion to find s:

v^2 = u^2 + 2as

(4.2 m/s)^2 = (0 m/s)^2 + 2 * (0.237 m/s^2) * s

17.64 m^2/s^2 = 0.474 m/s^2 * s

s = 17.64 m^2/s^2 / 0.474 m/s^2
s ≈ 37.2 m

Therefore, the length of the driveway is approximately 37.2 meters.