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?

Answer 1

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.