A 5.00-kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the bo

Question

A 5.00-kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

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Khang Minh 2 months 2021-08-28T07:56:18+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-28T07:57:31+00:00

    Explanation:

    According to the given situation,

           work done = change in kinetic energy

    As the formula of kinetic energy is as follows.

              K.E = \frac{1}{2}mv^{2}

    Putting the given values into the above formula as follows.

                 K.E = \frac{1}{2}mv^{2}

                       = \frac{1}{2} \times 5 kg \times (3 m/s)^{2}    

                       = 22.5 J

    Also we know that,

               Work done = force x distance

    or,          force = \frac{work}{distance}

                           = \frac{22.5 J}{4 m}

                           = 5.62 N

    In the given case, force is friction and formula to calculate friction is as follows.

             friction = \mu \times m \times g

    where, \mu = the coefficient of friction

    Putting the given values into the above formula as follows.

             Friction = \mu \times m \times g

             5.62N = \mu \times 5 \times 9.8  

             \mu = 0.114  

    Thus, we can conclude that the coefficient of kinetic friction between the floor and the given box is 0.114.

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