A driver in a $1000 - \mathrm { kg }$ car traveling at $24 \mathrm {~m} / \mathrm { s }$ slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the level road is 0.80, how long will the skid marks be?
A)37 m
B)30 m
C)46 m
D)34 m

Question

Quizplus · Accepted Answer

The work done by friction is equal to the initial kinetic energy of the car. Using the formula $ \text{friction force} = \mu \times m \times g $ and $ \text{work} = \text{friction force} \times \text{distance} $, we can solve for the distance. The initial kinetic energy is $ \frac{1}{2} m v^2 $. Solving gives the distance as 37 m.

A Driver in A 1000−kg1000 - \mathrm { kg }1000−kg Car Traveling At

A Driver in A $1000 - \mathrm { kg }$ Car Traveling At