Answer:
a) Em_f = ½ m vₐ², b) v = vₐ /√2
Explanation:
a) For this part we use conservation of energy
starting point .. block A moving, spring unstretched
Em₀ = K = ½ m vₐ²
end point. Stretched spring
Em_f = K_e = ½ k x²
energy is conserved
Em₀ = Em_f
Em_f = ½ m vₐ²
b) Let's analyze the movement a little, block A began to move at a speed va, I stretch the spring an amount Dx, it exerts a force on block b that begins to move and the elongation of the spring decreases.
In all this process there is no friction force, therefore the energy is conserved, therefore the maximum energy stored in the spring must be distributed among the bodies.
Em₀ = K_e = E₀
where E₀ is the initial energy of block a
E₀ = ½ m vₐ²
At the point where we are in equilibrium
Em_f = Kₐ + K_b = ½ m vₐ² +1/2 m [tex]v_{b}^2[/tex]
so that the spring does not stretch or shrink, the two bodies must go at the same speed.
Em_f = m v²
energy is conserved
Em₀ = Em_f
½ m vₐ² = m v²
v = vₐ /√2
therefore both blocks must go at this speed
