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2 November, 04:08

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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  1. 2 November, 05:39
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    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
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