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A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A.
What is the coefficient of kinetic friction, k , of the surface from A to B?

Respuesta :

Answer:

The coefficient of kinetic friction is 0.26

Explanation:

Given:

Mass of box [tex]m = 9[/tex] kg

Initial height [tex]h = 5[/tex] m

Final height [tex]h' = 0[/tex] m

Distance travel by block [tex]d = 19[/tex] m

For finding coefficient of kinetic friction,

We use conservation laws,

Work done by frictional force is equal to change in energy,

   [tex]W_{fr} = \Delta E[/tex]

Here only potential energy change so we can write,

 [tex]f_{k} d \cos 180 = mg(h') - mg(h)[/tex]

Here [tex]f_{k} = \mu_{k} mg[/tex]

[tex]-\mu_{k}\times mgd= -mgh[/tex]

  [tex]\mu_{k} = \frac{h}{d}[/tex]

  [tex]\mu _{k} = \frac{5}{19}[/tex]

  [tex]\mu _{k} = 0.26[/tex]

Therefore, the coefficient of kinetic friction is 0.26

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