A car moving at a steady 10 m/s on a level highway encounters a depression that has a circular cross-section with a radius of 30 m. The car maintains its speed as it drives through the depression.
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) 790 N
B) 200 N
C) 590 N
D) 390 N
E) 490 N

Question

Quizplus · Accepted Answer

At the bottom of the depression, the normal force is the sum of the gravitational force and the centripetal force required to keep the car moving in a circular path. The gravitational force (weight) is $mg = 60.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N}$. The centripetal force is given by $F_c = \frac{mv^2}{r}$, where $m = 60.0 \, \text{kg}$, $v = 10 \, \text{m/s}$, and $r = 30 \, \text{m}$. Thus, $F_c = \frac{60.0 \times 100}{30} = 200 \, \text{N}$. The total normal force is the sum of these two forces: $588 \, \text{N} + 200 \, \text{N} = 788 \, \text{N}$, which rounds to $790 \, \text{N}$.

A Car Moving at a Steady 10 M/s on a Level