A car passes a dip in the road, going first down, then up. At the very bottom of the dip, when the car's instantaneous velocity is passing through horizontal, how does the magnitude of the normal force on the car compare to the magnitude of the car's weight? A) $\left|\vec{F}_{N}\right|\left|\vec{F}_{g}\right|$ D) $\left|\vec{F}_{N}\right|\left|\vec{F}_{g}\right|$. E) We do not have enough information to answer.

Question

A car passes a dip in the road, going first down, then up. At the very bottom of the dip, when the car's instantaneous velocity is passing through horizontal, how does the magnitude of the normal force on the car compare to the magnitude of the car's weight?
A) $\left|\vec{F}_{N}\right|<\left|\vec{F}_{g}\right|$

B) $\left|\vec{F}_{N}\right|=\left|\vec{F}_{g}\right|$

C) $\left|\vec{F}_{N}\right|>\left|\vec{F}_{g}\right|$

D) $\left|\vec{F}_{N}\right|>\left|\vec{F}_{g}\right|$.

E) We do not have enough information to answer.

Quizplus · Accepted Answer

At the bottom of the dip, the car experiences centripetal acceleration in addition to gravity. The normal force must counteract both the weight of the car and provide the centripetal force needed to keep the car moving in a curved path. Therefore, the magnitude of the normal force is greater than the magnitude of the car's weight.

A Car Passes a Dip in the Road, Going First ∣F⃗N∣<∣F⃗g∣\left|\vec{F}_{N}\right|<\left|\vec{F}_{g}\right|​FN​​<​Fg​​

A Car Passes a Dip in the Road, Going First $\left|\vec{F}_{N}\right|<\left|\vec{F}_{g}\right|$