Answer:
A)The magnitude of the friction force is 1.0 × 10⁴ N.
B) It takes the car 4.1 s to come to stop.
C) On a wet road, the time and distance needed to stop the car increases by nearly 2.5 times. In this case, the car takes 10.4 s to stop and travels 145 m.
Step-by-step explanation:
Hi there!
A) The work done by friction to stop the car is equal to the change in kinetic energy of the car:
W = final kinetic energy - initial kinetic energy (final kinetic energy = 0 because the car stops)
W = - initial kinetic energy
The work is negative because it decreases the kinetic energy (KE) of the car.
Then:
W = KE
Fr · Δx = 1/2 · m · v²
Where:
Fr = friction force.
Δx = traveled distance.
m = mass of the car.
v = velocity of the car.
Solving the equation for Fr:
Fr = 1/2 · m · v² / Δx
Fr = 1/2 · 1500 kg · (27.8 m/s)² / 57 m
Fr = 1.0 × 10⁴ N
The magnitude of the friction force is 1.0 × 10⁴ N.
B) Let´s find the acceleration of the car using the equation of velocity and position:
v = v0 + a · t
x = x0 + v0 · t + 1/2 · a · t²
Where:
v = velocity.
v0 = initial velocity.
a = accleration.
t = time.
x = position at time t.
x0 = initial position.
We know that the car travels 57 m and that the final velocity is zero, so if t is the time it takes the car to stop:
0 = 27.8 m/s + a · t
57 m = 27.8 m/s · t + 1/2 · a · t²
We have a system of 2 equations with 2 unknowns:
Solving the first equation for "a":
a = -27.8 m/s / t
Replacing the acceleration in the equation of position:
57 m = 27.8 m/s · t + 1/2 · (-27.8 m/s / t) · t²
57 m = 27.8 m/s · t - 13.9 m/s · t
57 m = 13.9 m/s · t
t = 57 m / 13.9 m/s
t = 4.1 s
It takes the car 4.1 s to come to stop.
C) The friction force is calculated as follows:
Fr = μ · N
Where
μ = coefficient of friction.
N = normal force of the car.
If the coefficient of friction when the road is wet, μw, is 60% lower than μ, then:
μw = μ - 0.6 μ
and the friction force, Frw, will be:
Frw = μw · N = (μ - 0.6 μ) · N
Frw = μ · N - 0.6 μ · N
Frw = Fr - 0.6 Fr
Then, the friction force will also be 60% lower:
Frw = 1.0 × 10⁴ N - 0.6 · 1.0 × 10⁴ N = 4000 N
The sum of forces exerted on the car is equal to the friction force, so using Newton´s second law:
Frw = m · a
Solving for "a":
Frw/m = a
a = 4000 N / 1500 kg = 8/3 m/s²
Using the equation of velocity as before:
v = v0 + a · t
0 = 27.8 m/s - 8/3 m/s² · t (the acceleration is negative because the car is stopping)
Solving for t:
-27.8 m/s / -8/3 m/s² = t
t = 10.4 s
The time increases (10.4 s/4.1 s) nearly 2.5 times.
The distance traveled in that time will be:
x = x0 + v0 · t + 1/2 · a · t²
x = 27.8 m/s · 10.4 s - 1/2 · 8/3 m/s² · (10.4 s)²
x = 145 m
The distance increases (145 m / 57 m) 2.5 times as well.