) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kin

Question

) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface

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Nguyệt Ánh 3 years 2021-09-02T01:51:17+00:00 1 Answers 0 views 0

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  1. thachthao
    0
    2021-09-02T01:52:58+00:00
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    Answer:

    -0.3

    Explanation:

    F’ = μmg ……….. Equation 1

    Where F’ = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

    But,

    F’ = ma ………… Equation 2

    Where a = acceleration of the stone.

    Substitute equation 2 into equation 1

    ma = μmg

    dividing both side of the equation by m

    a = μg

    make μ the subject of the equation

    μ = a/g…………… Equation 3

    From the equation of motion,

    v² = u²+2as…………….. Equation 4

    Where v and u are the final and the initial velocity respectively, s = distance.

    Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

    Substitute into equation 4

    0² = 8² + 2×11×a

    22a = -64

    a = -64/22

    a = -32/11 m/s² = -2.91 m/s²

    substitute the values of a and g into equation 3

    μ = -2.91/9.8

    μ = -0.297

    μ ≈ -0.3

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