A 5.1 g marble is fired vertically upwardusing a spring gun. The spring must be

Question

A 5.1 g marble is fired vertically upwardusing a spring gun. The spring must be compressed 8.0 cm if themarble is to just reach a target 13 mabove the marble's position on the compressed spring. (a) What is the change in the gravitationalpotential energy of the marble-Earth system during the 13 m ascent?
1

(b) What is the change in the elastic potential energy of thespring during its launch of the marble?
2

(c) What is the spring constant of the spring?
3
(a) What is the change in the gravitationalpotential energy of the marble-Earth system during the 13 m ascent?
1

(b) What is the change in the elastic potential energy of thespring during its launch of the marble?
2

(c) What is the spring constant of the spring?
3

Explanation / Answer

The change in GPE = mgh PE = .0051kg*9.8*13m = 0.6497J The change in elastic potential energy needs to be equal tothe change in gravitational potential energy. Initially allthe marble had was the elastic potential energy from the spring andthat all gets converted to gravitational potential energy at thetop. Elastic Potential energy = 0.6497J Spring Constant: Elastic potential energy = 1/2kx2 k = 2*0.6497/.082 = 203.03N/m k = 2*0.6497/.082 = 203.03N/m

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