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9 March, 16:59

A 5.00-kg box slides 3.00 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

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  1. 9 March, 20:57
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    Let's call the coefficient of kinetic friction k.

    We can use the equation v^2 = v (initial) ^2 + 2a (x-x (initial)) to find the acceleration of the box.

    v=0 m/s

    v (initial) = 3 m/s

    x = 3 m

    x (initial) = 0 m

    a = (v^2-v (initial) ^2) / 2 (x-x (initial))

    a = (-9 m^2/s^2) / 3 m

    a = - 3 m/s^2

    Since F = ma, and the only horizontal force on the box is friction, we can say that F (friction) = k*F (normal) = k*F (gravitational) = m*a

    k = m*a/F (gravitational)

    F (gravitational) = m*g

    k = (m*a) / (m*g)

    k = a/g

    k = (-3 m/s^2) / (-9.81 m/s^2)

    k = 0.31
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