Question

A 5.2 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.0 cm if the marble is to just reach a target 13 m above the marble's position on the compressed spring.

(a) What is the change in the gravitational potential energy of the marble-Earth system during the 13 m ascent?
J

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

(c) What is the spring constant of the spring?
N/m

Answer 1

a) 0.66J

b) -0.66J

c) 206N/m

Step-by-step explanation:

The change of gravitational potencial energy is given by:

`E_g=m*g*h\\E_g=5.2*10^(-3)*9.8*13\\E_g=0.66J`

Because of energy conservation the elastic potencial energy was converted completely to gravitational so:

`E_e-E_g=0\\E_e=-E_g\\E_e=-0.66J`

The elastic potencial energy is given by:

`E_e=(1)/(2)k*x^2\\\\-0.66=-(1)/(2)K*(8*10^(-2))\\\\K=206N/m`