Chapter 4, Problem 42P

To determine

The initial speed of the car using the length of the skid marks and the coefficient of kinetic friction between rubber and the pavement.

Answer to Problem 42P

Solution:

The initial speed of the car is found to be 34 m/s.

Explanation of Solution

The car comes to rest due to the force of kinetic friction between the tyres of the car and the pavement. The frictional force provides a retardation, which brings the car to rest.

Given:

The distance the car skids before coming to rest $s=72\text{ m}$.

The coefficient of kinetic friction between rubber and the pavement ${\mu }_k=0.80$.

Formula used:

    $F_{fr}={\mu }_{k}mg$

    $F_{fr}=-ma$

    $v^2=v_0^2+2as$

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

Calculation:

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.

Since, $W=mg$, so 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 a retarding force on the car and brings it to rest under the action of a constant retardation a.

    $F_{fr}=-ma$. …………(2)

From equations (1)and (2),

    ${\mu }_{k}mg=-ma$

The retardation experienced by the car is given by,

    $a=-{\mu }_{k}g$…………(3)

Using the equation of motion, $v^2=v_0^2+2as$ and substituting 0 for v and acceleration from equation (3), write an expression for $v_0$.

    $\begin{array}{l}{v}^2=v_0^2+2as\end{array}$

    $\begin{array}{l}0=v_0^2+2\left(-{\mu }_{k}g\right)s\end{array}$

    $\begin{array}{l}{v}_0=\sqrt{2{\mu }_{k}gs}\end{array}$

Substitute the given values of s and ${\mu }_k$ in the above equation and calculate $v_0$.

    $\begin{array}{c}{v}_0=\sqrt{2{\mu }_{k}gs}\end{array}$

        $\begin{array}{c}=\sqrt{2\left(0.80\right)\left(9.8{\text{ m/s}}^2\right)\left(72\text{ m}\right)}\end{array}$

        $\begin{array}{c}=33.6\text{ m/s}\end{array}$