Question

1. the driver of a car made an emergency stop on a straight horizontal road. the wheels locked, and the car skidded to a stop. the marks made by the rubber tires on the dry asphalt are 16 meters long, and the car’s mass is 1200 kilograms. calculate the work done by the frictional force in stopping the car. coefficients of friction are found in the reference table.

Explanation:

Step1: Recall the formula for frictional force

The frictional force \( F_f \) on a horizontal surface is given by \( F_f=\mu_k F_N \), where \( \mu_k \) is the coefficient of kinetic friction and \( F_N \) is the normal force. On a horizontal road, the normal force \( F_N \) equals the weight of the car, so \( F_N = mg \), where \( m = 1200\space kg \) and \( g=9.8\space m/s^2 \). From reference tables, the coefficient of kinetic friction \( \mu_k \) for rubber on dry asphalt is approximately \( 0.85 \).

First, calculate the normal force:
\( F_N=mg = 1200\space kg\times9.8\space m/s^2= 11760\space N \)

Then, calculate the frictional force:
\( F_f=\mu_k F_N= 0.85\times11760\space N = 9996\space N \)

Step2: Recall the formula for work done by a force

The work done \( W \) by a force \( F \) over a distance \( d \) is given by \( W = Fd\cos\theta \). In this case, the frictional force acts opposite to the direction of motion, so \( \theta = 180^\circ \) and \( \cos\theta=- 1 \). The distance \( d = 16\space m \).

Substitute the values of \( F_f \) and \( d \) into the work formula:
\( W=F_f d\cos\theta=9996\space N\times16\space m\times(- 1) \)

But we are interested in the magnitude of the work done by friction to stop the car (or we can consider the work done against friction, but since friction is doing negative work to stop the car, the magnitude is what is asked in terms of the work done by friction to stop it).

\( W = - F_f d \) (the negative sign indicates that the force is opposite to the displacement). The magnitude of the work done by friction to stop the car is \( W = F_f d \) (since we are calculating the work done by friction in stopping the car, we can use the magnitude of the force and distance with the understanding of the direction).

\( W=9996\space N\times16\space m = 159936\space J\approx1.6\times 10^{5}\space J \) (or using more precise calculation)

If we use \( \mu_k = 0.8 \) (sometimes approximate values are used, but the standard value is around 0.8 - 0.9 for rubber on dry asphalt). Let's recalculate with \( \mu_k = 0.8 \):

\( F_N=1200\times9.8 = 11760\space N \)

\( F_f = 0.8\times11760=9408\space N \)

\( W=F_f\times d=9408\space N\times16\space m = 150528\space J\approx1.5\times 10^{5}\space J \)

(Note: The exact value depends on the precise value of \( \mu_k \). Using \( \mu_k = 0.85 \) gives \( W\approx1.6\times 10^{5}\space J \), using \( \mu_k = 0.8 \) gives \( W\approx1.5\times 10^{5}\space J \))

Answer:

The work done by the frictional force (using \( \mu_k = 0.85 \)) is approximately \( \boldsymbol{1.6\times 10^{5}\space J} \) (or with \( \mu_k = 0.8 \) approximately \( \boldsymbol{1.5\times 10^{5}\space J} \))