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After coming down a steep hill at a constant speed of 43 m/s, a car travels along the circumference of a vertical circle of radius 618 m until it begins to climb another hill. r x What is the magnitude of the net force on the 34 kg driver of the car at the lowest point on this circular path? Answer in units of kN.

User Dada
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1 Answer

7 votes

Answer:

F=0.101 kN

Step-by-step explanation:

Newton's 2nd law, F = ma, but this is circular path, the acceleration (a) is the centripetal acceleration.

a = (v²) / r

F = (m×v²) /r

F=(34 kg)×(43 m/s)² / 618 m

F=101.72 N

To convert Newtons into kilo-Newtons divide it 1000.

F=0.101 kN

User COMP Superscalar
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