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?

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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 $$v_{f}$$


1 Answer 1


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 $v_{f} =0$

In other word: When you are calculating things using energy conservation you always have to think about the initial and final states of the object.... And since you are looking for the highest point the box is able to reach, there is no kinetic energy in the final state, because if it would have a kinetic energy it would not be the highest point.

  • $\begingroup$ I understand now. so kf = 0 as well and all I have to do is solve for d2 $\endgroup$
    – MisterM
    Commented Nov 10, 2019 at 22:55
  • $\begingroup$ Yeah exactly. If this answers your question you may probably accept the anwser. physics.stackexchange.com/help/someone-answers $\endgroup$
    – B.Hueber
    Commented Nov 10, 2019 at 23:03
  • $\begingroup$ just did thank you $\endgroup$
    – MisterM
    Commented Nov 11, 2019 at 0:17

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