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22 December, 07:56

A 2.00-kg box slides on a rough, horizontal surface, hits a spring with a speed of 2.07 m/s, and compresses it a distance of 19.0 cm before coming to rest. Determine the force constant of the spring if the coefficient of kinetic friction between the block and the surface is μk = 0.660.

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  1. 22 December, 08:14
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    k = 101.2 N / m

    Explanation:

    For this exercise we can use the relationship between work and energy

    W = ΔK (1)

    Where the work of the friction force is

    W = fr x cos θ

    As the friction force opposes the movement, the angle is 180º, so the kinetic product remains

    W = - fr x

    The friction force is given by the equation

    fr = μ N

    Let's use Newton's second law

    Axis y

    N - W = 0

    N = W

    We substitute

    fr = μ mg

    So the work is

    W = - μ m g x

    On the other hand, the variation in energy is

    ΔEm = Em_final - Em_inicial

    ΔEm = ½ k x² - ½ m v²

    We substitute in our initial equation 1

    -μ m g x = ½ k x² - ½ m v²

    k = 2m / x² ( - μ g x + ½ v²)

    k = 2 2.00 / 0.190² ( - 0.660 9.8 0.190 + ½ 2.07²)

    k = 101.2 N / m
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