I have this problem in my physics homework. How far does the box travel along the ramp on its way up? [closed]

You are using a spring to launch 2kg boxes up a ramp. The spring has a spring constant of 500N/m and is compressed 50cm before launching each box from rest. You notice that there is a 1.5m long patch of the flat ground before the ramp that is sticky (assume all other surfaces are frictionless). The coefficient of friction between the box and the sticky patch is 1.2. How far does the box travel along the ramp on its way up?

What I know until now

Initial potential energy(Ui): $$\frac 12kx^2$$ Initial kinetic energy(Ki): $$0$$ Final potential energy(Uf): $$mgd_{2}sin40^\circ$$ Final kinetic energy(Kf): $$\frac 12mv_{f}^2$$ Work done by non conservative forces (friction for sticky patch)(Wnc): $$-\mu mgd_{1}$$

and I know the conservation of energy formula here is: $$U_{i} + K_{i} + W_{nc} = U_{f}+K_{f}$$

I could solve for $$d_{2}$$ but I do not know how to find the final velocity $$v_{f}$$

There is no final velocity, because the final point is exactly the point where the box stands still on the ramp and afterwards begins to slide the way down again... So $$v_{f} =0$$