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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?

User Ali Navidi
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1 Answer

6 votes

Answer:

L = 2.98 m

Step-by-step 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


Em_(f) = 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

User Scottsanpedro
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