A1500 kg car driving at 100 km/hr brakes and comes to a stop after sliding for 57 m.

a) calculate the magnitude of the friction force between the car and road.
b) calculate the time the car takes to come to a stop.
c) a wet road typically has a coefficient of friction 60% lower than that of a dry road.
how would this change the distance and time it takes the car to stop

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.

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.

6/12=1/2

step-by-step explanation:

correct it is c.) b b