A spring is compressed by 50 cm from it's equilibrium length by an unattached block of mass 5.0 kg (the spring constant is 400 N/m). The box is released from rest, slides along a friction-less surface for a time, and then ascends a ramp of length 10 m, which is inclined at an angle of 20 degrees above the horizontal. What is the maximum distance that the box travels up the ramp?

  L = 2.98 m

Explanation:

In this exercise we are told that there is no friction, so we can use energy conservation

Starting point. Compressed spring

        Em₀ = Ke = ½ k x²

Final point. At the highest point of the ramp

        = U = mg h

As there is no friction the energy is conserved

         Emo = Em_{f}

         ½ k x² = mg h

         h = ½ k x² / mg

         h = ½ 400 0.50² / (5.0 9.8)

         h = 1.02 m

This is the height that the body reaches, to calculate the distance traveled on the ramp let's use trigonometry

          sin θ = h / L

         L = h / sin θ

         L = 1.02 / sin 20

         L = 2.98 m

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object

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