A 1200 kg car’s tires have a coefficient of kinetic friction of 0.88 with the road surface. While the car is driving

Question

along, the engine stalls, the brakes seize up, and friction causes the car to eventually slide to a stop.
a) Draw a Free-Body Diagram for the car.
b) What is the force of gravity on the car?
c) What is the normal force on the car?
d) What is the magnitude of the force of friction on the car?
e) What is the acceleration of the car?

Answer 1

Let's break this down step by step.

a) Free-Body Diagram (FBD)

To draw a Free-Body Diagram for the car, we identify the forces acting on it:

• Weight (Gravitational force): downward, represented as $F_g$
• Normal Force: upward, represented as $F_n$
• Frictional Force: directed opposite to the direction of motion (if the car is sliding to the right, friction acts to the left), represented as $F_f$

In practice, the FBD would look like this:

         ↑
        F_n
         |
 -----------------
|        Car      |
 -----------------
         |
         ↓
        F_g
        
        ←
        F_f

b) Force of Gravity on the Car

The force of gravity can be calculated using the formula: $$F_g=m\cdot g$$ where:

• $m=1200,\text{kg}$ (mass of the car),
• $g=9.81,{\text{m/s}}^2$ (acceleration due to gravity).

So, $$F_g=1200,\text{kg}\cdot 9.81,{\text{m/s}}^2=11772,\text{N}$$

c) Normal Force on the Car

Since the car is on a flat surface and there are no vertical accelerations, the normal force $F_n$ will be equal to the force of gravity: $$F_n=F_g=11772,\text{N}$$

d) Magnitude of the Force of Friction on the Car

The force of friction $F_f$ can be calculated using the formula: $$F_f={\mu }_k\cdot F_n$$ where:

• ${\mu }_k=0.88$ (coefficient of kinetic friction),
• $F_n=11772,\text{N}$.

So, $$F_f=0.88\cdot 11772,\text{N}=10335.36,\text{N}$$

e) Acceleration of the Car

The acceleration can be found using Newton's second law: $$F=m\cdot a$$ Here, the only horizontal force acting is the frictional force, which acts to decelerate the car. Thus, we have: $$F_f=m\cdot a$$ Rearranging gives: $$a=\frac{F_f}{m}$$ Now, substituting the values we found: $$a=\frac{10335.36,\text{N}}{1200,\text{kg}}\approx 8.61,{\text{m/s}}^2$$

Since the friction force opposes the motion of the car, the acceleration will be negative: $$a\approx -8.61,{\text{m/s}}^2$$

Summary of Answers:

a) Free-Body Diagram is described above.
b) $F_g=11772,\text{N}$
c) $F_n=11772,\text{N}$
d) $F_f\approx 10335.36,\text{N}$
e) $a\approx -8.61,{\text{m/s}}^2$