A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the bottom of the depression?
A) 200 N
B) 389 N
C) 789 N
D) 489 N
E) 589 N

Question

Quizplus · Accepted Answer

When the car is at the bottom of the depression, the net force acting on the passenger will be the centripetal force that keeps the passenger moving in a circular path.

Centripetal force = m(v^2/r)

where m is the mass of the passenger, v is the speed of the car, and r is the radius of the depression.

Plugging in the given values, we get:

Centripetal force = (60.0 kg)(10.0 m/s)^2 / 30.0 m = 200 N

The normal force exerted by the seat on the passenger is equal in magnitude to the centripetal force, since the passenger is also subject to gravity.

Normal force = Centripetal force = 200 N

Therefore, the best choice is (C), 789 N.

A Car Moving at 10