A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in motion, in

Question

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in motion, in horizontal force of 650N

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Philomena 3 years 2021-07-23T00:44:47+00:00 1 Answers 699 views 0

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  1. Euphemia
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    2021-07-23T00:46:44+00:00
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    Complete Question

    A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

    Answer:

    The value for static friction is \mu_s = 0.60

    The value for static friction is \mu_k = 0.70

    Explanation:

    From the question we are told that

    The mass of the clock is m = 95 \ kg

    The first horizontal force is F _s = 560 \ N

        The second horizontal force is    F _k = 650 \ N

    Generally the static frictional force is equal to the first  horizontal force

    So

         F _s = m * g * \mu_s

    =>   560 = 95 * 9.8 * \mu_s

    =>    \mu_s = 0.60

    Generally the kinetic frictional force is equal to the second horizontal force

    So

          F _k = m * g * \mu_k

          650 = 95 * 9.8 * \mu_k

         \mu_k = 0.70

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