(8c8p15) A 5.5 g marble is fired vertically upward using a spring gun. The sprin

Question

(8c8p15) A 5.5 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.5 cm if the marble is to just reach a target 13 m above the marble's position on the compressed spring. What is the change in the gravitational potential energy of the marble-Earth system during the 13 m ascent? Submit Answer Tries 0/8 What is the change in the elastic potential energy of the spring during its launch of the marble? Submit Answer Tries 0/8 What is the spring constant of the spring?

Explanation / Answer

<< What is the change in the gravitational potential energy of the marble-Earth system during the 13 m ascent? >>

PE = mgh

PE = (5.5/1000)* 9.8 * 13

PE = 0.700 J


<< What is the change in the elastic potential energy of the spring during its launch of the marble? >>

Using the the conservation of energy, the energy absorbed in the marble should be equal to the energy from the spring. Hence,

Spring energy = Potential energy of marble = 0.700 J


<< What is the spring constant of the spring? >>

Spring energy = (1/2)kx^2

0.700 = (1/2)(k)(8.5/100)^2

Solving for "k"

k = 194 N/m

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