a small car with mass .800kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00m. if the normal force exerted by the track (point b) is 6.00n, what is the normal force on the car when it is at the bottom of the track (point a)?

Question

08-11-2022
Physics

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a small car with mass .800kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00m. if the normal force exerted by the track (point b) is 6.00n, what is the normal force on the car when it is at the bottom of the track (point a)?

Answer :

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kjuli1766 kjuli1766 · Answer 1 · 2022-11-11T09:21:18-05:00

The normal force on the car when it is at the bottom of the track is 22 N.

A frame revolves in a vertical circle such that its motion at unique points is one-of-a-kind then the motion of the frame is said to be vertical circular movement. recollect an item of mass tied at one give up of a mass m tied at one stop of an int extensible string and whirled in a vertical radius of radius r.

Calculation:-

At the inside of the track

N + mg = mv²/r

6 + 0.8 × 10 = m (v²/r)

14 = 0.8 ( acceleration)

acceleration = 17.5 m/sec² towards centre.

Again,

a = v²/r

ar = v²

v = √ar

v = √17.5 ×5

= 9.35 m/s²

For Normal force

N = mv²/r + mg

N = 0.8 × 1.75 + 0.8 × 10

= 14 + 8

= 22 N

A frame spins in a vertical circle in such a way that its movement at exclusive factors differs from the frame's motion, which is assumed to be vertical circular motion. keep in mind a mass m item wrapped in a stretchable string at one end and whirled in a vertical circle with radius d.

Learn more about vertical circles here:-https://brainly.com/question/19904705

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