Physics question

Question

A car has a mass of 1020 kg and has a coefficient of static friction between the ground and its tires of 0.85. What force of friction can it exert on the ground? What is the maximum acceleration of this car? If its initial speed was 27 m/s, how far does it travel before stopping?

Answer 1

Hi Ashy!

The problem has three parts.

1) What is the maximum force of friction the car can exert on the ground?

Maybe you remember from physics class, but the maximum force of static friction (call it f_s,max) that an object can exert on the ground is μ_sN (where N is the normal force exerted by the ground on the car and μ_s is the coefficient of static friction.)

In equation form, f_s,max = μ_sN

We know μ_s, so now we just have to find N, the normal force. For this part it's useful to draw a free body diagram. There are only three forces acting on the car. Gravity, the Normal force, and Friction. Gravity points straight down, the Normal force points straight up, and friction points in the direction opposite to the car's motion.

Now, because the car is not accelerating upwards or downwards, we know that the normal force is exactly equal to the force of gravity, so N = mg.

Going back to our first equation, we have now that f_s,max = μ_smg = (0.85)(1020 kg)(9.8 m/s²) = 8496.6 N.

And that's our answer!

2) What is the maximum acceleration of this car?

For this next part, we want to find the maximum acceleration of this car. Well, what force is going to be able to accelerate the car? The answer is it's the frictional force that the ground exerts on the car!

Do you remember Newton's second law F = ma? Well this tells us that the the acceleration of an object is proportional to the net force exerted on it. We know that F in this case is just f_s,max that we calculated in part 1.

f_s,max = μ_smg = ma_car,max

μ_smg = ma_car,max

We can cancel the m on both sides, which shows us that

a_car,max = μ_sg = (0.85)(9.8 m/s²) = 8.33 m/s²

3) If the car's initial speed was 27 m/s, how far does it travel before stopping?

Now that we have all that, we are now asked how far the car travels before stopping if it had an initial speed of 27 m/s. Well, we know that the car is going to slow down because of the frictional force that we calculated, right? And we actually already calculated the acceleration that the car experiences due to the frictional force in part 2. It's 8.33 m/s^2.

Have you seen this equation before in your physics class?

v_xf² = v_x0² + 2a_xd

v_xf is the car's final speed

v_x0 is the car's initial speed

a_x is the acceleration of the car

d is the distance traveled during the deceleration.

If we think about our scenario, the car starts at a velocity of 27 m/s, so v_x0 is 27 m/s. At the end, the car is stopped, so its final velocity is zero, meaning that v_xf = 0. We know that the acceleration of the car is negative, because the car is decelerating, not accelerating, so a_x = -8.33 m/s². And d is what the problem is asking us for!

If we plug these values in, we get that

v_xf² = v_x0² + 2a_xd

0 = (27 m/s)² + 2(-8.33 m/s²)(d)

Rearranging, we get that

16.66d = (27)²

d = (27)²/16.66 = 43.76 m

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I hope this helped! If you would like a better explanation or need help with other questions in the future, feel free to contact me!

-- Alex