A 2.1 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an average friction force of 4.0 103 N impedes the motion, find the speed of the car at the bottom of the driveway.
PE=KE+W(fr)
PE=m•g•h=m•g•s•sinα,
KE = mv²/2,
W(fr)=F(fr) •s•sinα.
m•g•s•sinα=mv²/2+F(fr) •s•sinα.
Solve for ‘v’
To find the speed of the car at the bottom of the driveway, we can use the principle of conservation of energy. The total mechanical energy of the system (car and Earth) is conserved if we neglect air resistance.
First, let's find the potential energy (PE) of the car at the top of the driveway. The potential energy is given by the formula:
PE = m * g * h,
where m is the mass of the car, g is the acceleration due to gravity, and h is the height.
Given:
m = 2.1 * 10^3 kg
g = 9.8 m/s^2
h = 5.9 m
Substituting these values into the formula, we get:
PE = (2.1 * 10^3 kg) * (9.8 m/s^2) * (5.9 m)
= 1.23 * 10^5 Joules
Next, let's consider the kinetic energy (KE) of the car at the bottom of the driveway. The kinetic energy is given by the formula:
KE = (1/2) * m * v^2,
where v is the velocity of the car.
Now, considering the work done by friction, we can find the change in mechanical energy of the system. The work done by friction is given by the formula:
Work = force * distance * cos(theta),
where force is the average friction force, distance is the length of the driveway, and theta is the angle between the force and the displacement.
Given:
force = 4.0 * 10^3 N
distance = 5.9 m
theta = 19°
Substituting these values into the formula, we get:
Work = (4.0 * 10^3 N) * (5.9 m) * cos(19°)
≈ 3.97 * 10^4 Joules
Since work is equal to the change in mechanical energy, we can set up the equation:
KE + PE = Work
Substituting the values we obtained, we have:
(1/2) * m * v^2 + m * g * h = Work
Simplifying and solving for v^2, we get:
(1/2) * v^2 = Work - m * g * h
v^2 = 2 * (Work - m * g * h) / m
v = √(2 * (Work - m * g * h) / m)
Substituting the known values, we have:
v = √(2 * (3.97 * 10^4 J - (2.1 * 10^3 kg) * (9.8 m/s^2) * (5.9 m)) / (2.1 * 10^3 kg))
v ≈ 13.2 m/s
Therefore, the speed of the car at the bottom of the driveway is approximately 13.2 m/s.