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11 May, 10:48

An object slides on a level surface in the + x direction. It slows and comes to a stop with a constant acceleration of - 2.45 m/s2. What is the coefficient of kinetic friction between the object and the floor?

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  1. 11 May, 10:58
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    uk = 0.25

    Explanation:

    Given:-

    - An object comes to stop with acceleration, a = - 2.45 m/s^2

    Find:-

    What is the coefficient of kinetic friction between the object and the floor?

    Solution:-

    - Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).

    - We will apply equilibrium equation on the object in vertical direction.

    N - m*g = 0

    N = m*g

    Where, N : Contact force exerted by the surface on the floor

    g : Gravitational acceleration constant = 9.81 m/s^2

    - Now apply Newton's second law of motion in the horizontal (x-direction):

    - Ff = m*a

    - The frictional force is related to contact force (N) by the following expression:

    Ff = uk*N

    - Substitute the 1st and 3rd expressions in the 2nd equation:

    uk*m*g = - m*a

    uk = a / g

    - Plug in the values and solve for uk:

    uk = - (-2.45) / 9.81

    uk = 0.25
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