## The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of this portion o

Question

The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of this portion of the track is 57.5 cm, what are the magnitude and direction of the force the track exerts on the car

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2 weeks 2021-08-30T21:09:40+00:00 1 Answers 0 views 0

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r………………………….. Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track