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Sunny_sXe [5.5K]
3 years ago
8

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, introd

uced 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?
Physics
1 answer:
NARA [144]3 years ago
3 0

Answer:

The coefficient of kinetic friction is 0.26

Explanation:

Given:

Mass of box m = 9 kg

Initial height h = 5 m

Final height h' = 0 m

Distance travel by block d = 19 m

For finding coefficient of kinetic friction,

We use conservation laws,

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

   W_{fr} = \Delta E

Here only potential energy change so we can write,

 f_{k} d \cos 180 = mg(h') - mg(h)

Here f_{k} = \mu_{k} mg

-\mu_{k}\times mgd= -mgh

  \mu_{k} = \frac{h}{d}

  \mu _{k} = \frac{5}{19}

  \mu _{k} = 0.26

Therefore, the coefficient of kinetic friction is 0.26

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