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A 5.00-kg box slides 5.50 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?

Respuesta :

cjrodriguez89

Answer:

μk = 0.08

Explanation:

  • Applying the work-energy theorem, we know that the change in kinetic energy of the box, is equal to the total work done on it.
  • The only force acting on the box capable of doing work, is the dynamic friction force.
  • This force is just the product of  the coefficient of kinetic friction, and the normal force.
  • In this case, if the floor is level, the normal force is equal and opposite to the gravity force, Fg.
  • So we can write the following expression:

       [tex]\Delta K = -\frac{1}{2} * m* v_{0} ^{2} = \mu_{k} *m*g* cos (180\º)[/tex]

  • Replacing by the givens, we can solve for μk, as follows:

       [tex]\mu_{k} = \frac{v_{0}^{2} }{2*g*d} = \frac{9 (m/s)2}{2*9.8 m/s2*5.50m} = 0.08[/tex]

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