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