A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m.What is the magnitude of the average stopping force exerted on the car?
A)690 N
B)550 N
C)410 N
D)340 N

Question

Quizplus · Accepted Answer

The work done to stop the car is equal to the change in kinetic energy, which can be calculated using the formula $ \Delta KE = \frac{1}{2}mv^2 $, where $ m $ is the mass of the car and $ v $ is its initial velocity. Plugging in the given values, $ \Delta KE = \frac{1}{2} \times 1100 \times 27^2 = 400950 $ Joules. The work done is also equal to the force times the distance over which it is applied, $ W = F \times d $. Rearranging for $ F $, we get $ F = \frac{W}{d} = \frac{400950}{578} \approx 693.6 $ N, which rounds to 690 N, indicating the magnitude of the average stopping force.

A Car of Mass 1100 Kg That Is Traveling at 27