Question

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reach a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

Answer 1

a)
`\Delta U_(g) = 12.945\,J`, b)
`\Delta U_(k) = 12.945\,J`, c)
`k = 2930.059\,(N)/(m)`

Step-by-step explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

`\Delta U_(g) = (0.06\,kg)\cdot \left(9.807\,(m)/(s^(2))\right)\cdot (22\,m)`

`\Delta U_(g) = 12.945\,J`

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

`\Delta U_(k) = 12.945\,J`

c) The spring constant of the gun is:

`\Delta U_(k) = (1)/(2) \cdot k \cdot x^(2)`

`k = (2\cdot \Delta U_(k))/(x^(2))`

`k = (2\cdot (12.945\,J))/((0.094\,m)^(2))`

`k = 2930.059\,(N)/(m)`