The sum of the forces exerted on a 1000-kg car while it is slowing down from 30 m/s to 15.9 m/s is 9500 N.How far does it travel while slowing down?
A)34 m
B)31 m
C)37 m
D)41 m

Question

Quizplus · Accepted Answer

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. The work done by the forces slowing down the car can be calculated using the formula $W = F \cdot d$, where $W$ is the work done, $F$ is the force, and $d$ is the distance over which the force is applied. The change in kinetic energy ($\Delta KE$) can be calculated using the formula $\Delta KE = \frac{1}{2} m (v_f^2 - v_i^2)$, where $m$ is the mass of the car, $v_f$ is the final velocity, and $v_i$ is the initial velocity. Setting the work done equal to the change in kinetic energy gives us:$$9500 \cdot d = \frac{1}{2} \cdot 1000 \cdot (15.9^2 - 30^2)$$Solving for $d$ gives:$$d = \frac{\frac{1}{2} \cdot 1000 \cdot (15.9^2 - 30^2)}{9500}$$$$d \approx 34 \, \text{m}$$Therefore, the car travels approximately 34 meters while slowing down.

The Sum of the Forces Exerted on a 1000-Kg Car