A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to $15.9 \mathrm {~m} / \mathrm { s }$ 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 done by the net force on the car is equal to the change in kinetic energy of the car. The work done by the net force is:$$W = Fd = 9500 N 	imes d$$The change in kinetic energy of the car is:$$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}(1000 kg)(15.9 m/s)^2 - \frac{1}{2}(1000 kg)(30 m/s)^2 = -220500 J$$The negative sign indicates that the kinetic energy of the car is decreasing.Therefore, the work done by the net force is equal to the change in kinetic energy of the car:$$Fd = \Delta K$$$$9500 N 	imes d = -220500 J$$$$d = \frac{-220500 J}{9500 N} = -23.2 m$$The negative sign indicates that the car is traveling in the opposite direction of the net force.Therefore, the car travels 23.2 m while slowing down.

A 1000-Kg Car Experiences a Net Force of 9500 N 15.9 m/s15.9 \mathrm {~m} / \mathrm { s }15.9 m/s

A 1000-Kg Car Experiences a Net Force of 9500 N $15.9 \mathrm {~m} / \mathrm { s }$