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 on a horizontal surface

The frictional force \( F_f \) on a horizontal surface is given by \( F_f=\mu N \), where \( N \) is the normal force. On a horizontal surface, \( N = mg \) (since the normal force balances the weight of the car), so \( F_f=\mu mg \). Here, \( m = 1200\space kg \), \( g = 9.8\space m/s^2 \), and we need the coefficient of kinetic friction \( \mu \) for rubber on dry asphalt. From reference tables, \( \mu = 0.8 \) (for rubber on dry asphalt, kinetic friction coefficient).

Step2: Calculate the frictional force

Substitute the values into the frictional force formula:
\( F_f=\mu mg = 0.8\times1200\times9.8 \)
\( F_f = 0.8\times11760 = 9408\space N \)

Step3: Recall the formula for work done by a force

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

Step4: Calculate the work done by friction

Substitute \( F_f = 9408\space N \), \( d = 16\space m \), and \( \cos\theta=-1 \) into the work formula:
\( W = F_f\times d\times\cos\theta=9408\times16\times(- 1) \)
But we are interested in the magnitude of the work done to stop the car (or the work done by friction, considering the direction). Alternatively, since friction opposes motion, the work done by friction is negative of the work done to stop, but let's compute it:
\( W=-9408\times16=- 150528\space J \). The magnitude of the work done by friction to stop the car is \( 150528\space J \) (or we can say the work done by friction is \( - 1.5\times10^5\space J \) approximately, but using exact calculation: \( 0.8\times1200\times9.8\times16\times(- 1)=-150528\space J \)).

Answer:

The work done by the frictional force is \(\boldsymbol{- 150528\space J}\) (or approximately \(-1.5\times 10^{5}\space J\)). If we consider the magnitude (work done to stop the car), it is \(150528\space J\).