Un cubo de masa m = 4,00 kg está unido a un resorte de constante k = 600 N/m como se muestra en la figura. El bloque se jala hasta una posición xi = 5,00 cm a la derecha del equilibrio y se suelta del reposo. Encuentra la velocidad que tiene el bloque al pasar por el equilibrio si a) la superficie horizontal no tiene fricción y b) el coeficiente de fricción entre el bloque y la superficie es µk=0,350

a)   v = 0.612 m / s , b) v = 0.179 m / s

Explanation:

We can solve this problem using the concepts of conservation of mechanical energy

starting point. Before releasing the cube

          Em₀ = = ½ k x²

final point. At the equilibrium point of the spring

          = K = ½ m v²

a) in this case there is no friction, so the energy is conserved

          Emo = Em_{f}

          ½ k x² = ½ m v²

          v = √k/m    x

let's reduce the magnitudes to the SI system

   x = 5.00 cm = 0.0500 m

let's calculate

         v = √600/4    0.05

         v = 0.612 m / s

b) In this part we are told that there is a friction, for which we use the work energy theorem

            W = ΔEm

friction force work is

           W = fr. x = fr x cos θ

the friction force opposes the movement, so the angle is 180º and the cosine is -1

the formula for friction force is

         fr = μ N

let's write Newton's second law on the vertical axis

         N - W = 0

         N = W

we substitute the friction force

         fr = μ m g

of the work energy equation

       - fr x = Em_{f} - Em₀

        - μ m g x = ½ m v² - ½ k x²

          v² = (½ k x² - μ m g x) 2 / m

          v = √(k / m x² - 2 μ g x)

          v = √ (600/4  0.05² - 2 0.350 9.8 0.05)

          v = 0.179 m / s

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

so easy. just google the electromagnetic waves and it’ll pop up.