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?

Explanation:

According to the given situation,

       work done = change in kinetic energy

As the formula of kinetic energy is as follows.

          K.E =

Putting the given values into the above formula as follows.

             K.E =

                   =    

                   = 22.5 J

Also we know that,

           Work done = force x distance

or,          force =

                       =

                       = 5.62 N

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

         friction =

where, = the coefficient of friction

Putting the given values into the above formula as follows.

         Friction =

         5.62N =  

          = 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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