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 poin

Question

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?

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Thu Thủy 2 months 2021-07-29T13:31:13+00:00 1 Answers 20 views 0

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    2021-07-29T13:32:54+00:00

    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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