A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to 15.9 m/s. How far does it travel while slowing down?
A)34 m
B)31 m
C)37 m
D)41 m

Question

A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to 15.9 m/s.  How far does it travel while slowing down?
A)34 m
B)31 m
C)37 m
D)41 m

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 given by:$$W = Fd$$where:* W is the work done (in joules)* F is the net force (in newtons)* d is the distance traveled (in meters)The change in kinetic energy of the car is given by:$$KE = \frac{1}{2}mv^2$$where:* KE is the kinetic energy (in joules)* m is the mass of the car (in kilograms)* v is the velocity of the car (in meters per second)Setting the work done by the net force equal to the change in kinetic energy, we get:$$Fd = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i$$where:* v_f is the final velocity of the car (in meters per second)* v_i is the initial velocity of the car (in meters per second)Substituting the given values into the equation, we get:$$9500 d = \frac{1}{2}(1000)(15.9)^2 - \frac{1}{2}(1000)(30)^2$$$$9500 d = -115,550$$$$d = -12.16 m$$The negative sign indicates that the car is traveling in the opposite direction of the net force. Therefore, the distance traveled by the car while slowing down is 12.16 m.

A 1000-Kg Car Experiences a Net Force of 9500 N