Chapter 4, Problem 43P

To determine

The coefficient of kinetic friction between the drag-race tires in contact with an asphalt surface.

Answer to Problem 43P

Solution:

The coefficient of kinetic friction between the drag-race tires and asphalt was found to be 1.4.

Explanation of Solution

During a drag-race, the vehicle exerts a very strong force on the ground. This in turn exerts a very large acceleration on the vehicle. The force of kinetic friction between the tires and the surface provides the acceleration for the vehicle.

Given:

The distance covered by the drag-racer $s=1\text{ km}=1000\text{ m}$

The time taken to cover the distance $t=12\text{ s}$

The initial speed of the drag-racer is $v_0=0$

Formula used:

Frictional force can be obtained by:

  $F_{fr}={\mu }_k{F}_N$

According to Newton’s second law of motion:

  $F_{fr}=ma$

Second equation of motion is:

  $s=v_{0}t+\frac{1}{2}a{t}^2$

Here, the force of kinetic friction is $F_{fr}$ , acceleration due to gravity is g , the mass of the car is m , its acceleration is a , initial speed is $v_0$ , its final speed is v and the time taken to travel this distance is t .

The weight W of the car acts vertically downwards and due to Newton’s third law, the car experiences an upward force called normal force $F_N$ .Since, there is no motion either in the upward and the downward directions,

  $F_N+W=0$

Therefore,

  $F_N=-W$

The normal force has the same magnitude as that of the weight and it acts opposite to the direction of the weight.

As the magnitude of the normal force is $F_N=mg$ .

The force of kinetic friction is related to the normal force as,

  $F_{fr}={\mu }_k{F}_N$

Therefore $F_{fr}={\mu }_{k}mg$   ......(1)

This force provides accelerationa on the vehicle.

  $F_{fr}=ma$   .......(2)

From equations (1) and (2),

  ${\mu }_{k}mg=ma$

The acceleration experienced by the car is given by,

  $a={\mu }_{k}g$   ......(3)

Calculation:

The acceleration of the vehicle can be calculated using the equation of motion,

  $s=v_{0}t+\frac{1}{2}a{t}^2$

Since, the vehicle starts from rest, $v_0=0$ .

Therefore,

  $\begin{array}{c}s=\frac{1}{2}a{t}^2\\ a=\frac{2s}{t^2}\end{array}$   ......(4)

From equations (3) and (4),

  $\begin{array}{c}a=\frac{2s}{t^2}={\mu }_{k}g\\ {\mu }_k=\frac{2s}{g{t}^2}\end{array}$

Using the given values of s , g and t in the equation

  $\begin{array}{c}{\mu }_k=\frac{2s}{g{t}^2}\\ =\frac{2\left(1000\text{ m}\right)}{\left(9.8{\text{ m/s}}^2\right){\left(12\text{ s}\right)}^2}\\ =1.417\end{array}$

Conclusion:

The drag-racer would cover a distance of 1 km in 12 s starting from rest if the coefficient of kinetic friction between the tires and the asphalt surface is 1.4.