Q31E - University Physics with Modern Physics

Question

A car is traveling on a level road with speed ${\mathit{v}}_{\mathbf{0}}$ at the instant when the brakes lock, so that the tires slide rather than roll.

(a) Use the work–energy theorem to calculate the minimum stopping distance of the car in terms of ${\mathit{v}}_{\mathbf{0}}$ , g, and the coefficient of kinetic friction ${\mathit{\mu }}_{\mathbf{k}}$ between the tires and the road. 

(b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

Step-by-Step Solution

Answer

a) The minimum stopping distance of the car is $\frac{v_0^2}{2{\mu }_{k}gs}$ .

b) (i) The minimum stopping distance is half when the coefficient of the kinetic friction is doubled.

(ii) The value of the minimum stopping distance is four times when the initial velocity was doubled.

(iii) The value of the minimum stopping distance is double when the coefficient of the kinetic friction doubles and the initial velocity is doubled.

Step 1: Identification of given data

The given data can be listed below,

• The initial speed of the car is, $v_0$
• The coefficient of the kinetic friction is, ${\mu }_k$

Step 2: Concept/Significance of coefficient of kinetic friction.

The ratio of kinetic friction force to the normal force applied to a body is known as the coefficient of kinetic friction.

Step 3: (a) Determination of the minimum stopping distance of the car in terms of v 0

The work done on the car is given by,

 

$$\begin{gathered} W_{to t}=K_f-K_i \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_j^2 \end{gathered}$$                                                                                                        …(i)

 

Here, m is the mass of the car, v is the initial velocity of the car and $v_f$ is the final velocity of the car.

 

Also, the work done due to frictional force is given by,

 

$W_{to t}={\mu }_{k}mgs$                                                                                                                       …(ii)

 

Here ${\mu }_k$ is the coefficient of kinetic friction, g is the acceleration due to gravity, and s is the distance traveled by car.

 

Compare equation (i) and (ii), the minimum distance of the car is given by,

$$\begin{gathered} {\mu }_{k}mgs=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_j^2 \\ -{\mu }_{k}mgs=0-\frac{1}{2}m{v}_0^2 \\ s=\frac{v_0^2}{2{\mu }_{k}gs} \end{gathered}$$

Thus, the minimum stopping distance of the car is $\frac{v_0^2}{2{\mu }_{k}gs}$ .

Step 4: (b) Determination factor would the minimum stopping distance change (i) The coefficient of kinetic friction was doubled

The stopping distance is given by,

 

$s=\frac{v_0^2}{2{\mu }_{k}gs}$

 

Substitute the value of the coefficient of kinetic friction in the above,

 

$$\begin{gathered} s_1=\frac{v_0^2}{2\times 2{\mu }_{k}gs} \\ =\frac{1}{2}s \end{gathered}$$

 

Thus, the minimum stopping distance is half when the coefficient of the kinetic friction is halved.

Step 5: (ii) The initial speed was doubled

The stopping distance is given by,

$s=\frac{v_0^2}{2{\mu }_{k}gs}$

 

Substitute the value of the coefficient of kinetic friction in the above,

 

$$\begin{gathered} s_1=\frac{{\left(2v_0\right)}^2}{2{\mu }_{k}gs} \\ =4s \end{gathered}$$

 

Thus, the value of the minimum stopping distance is four times when the initial velocity was doubled.

Step 6: (ii) The coefficient of kinetic friction and the initial speed were doubled

The stopping distance is given by,

 

$s=\frac{v_0^2}{2{\mu }_{k}gs}$

 

Substitute the value of the coefficient of kinetic friction in the above,

 

$$\begin{gathered} s_1=\frac{v_0^2}{2\times 2{\mu }_{k}gs} \\ =2s \end{gathered}$$

 

Thus, the value of the minimum stopping distance is double when the coefficient of the kinetic friction is double, and the initial velocity is doubled.