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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?



Answer :

The automobile, which weighs 388 newtons, is at the bottom of the depression. At 17.15 m/s, the automobile loses contact with the bumpy road.

The normal force is the force that surfaces use to prevent solid objects from piercing one another. Normal forces include contact forces. Two surfaces that are not in contact cannot be moved by a normal force.

Calculation:

r = 30 m

v = 10 m/s

m = 60 kg

(a) Let N be the normal reaction.

At the bump

N = mg - mv^2 / r

N = 60 x 9.8 - 60 x 10 x 10 / 30 = 588 - 200 = 388 newton

(b) Then the contact loose, N = 0

So, mg = mv^2 / r

v^2 = r x g = 30 x 9.8 = 294

v = 17.15 m/s.

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