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.

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

if they have a finite resultant

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