Question

A 5.0 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 20 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 20 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

Answer:

a. 0.98 J

b. -0.98 J

c. 306.25 J/m²

Explanation:

(a)

$\Delta U_g = mgh\\\Delta U_g = 5\times 10^{-3}\times 9.8 \times 20\\\Delta U_g = 0.980 J$

(b)

The increase in gravitational potential energy = decrease in elastic potential energy

Elastic potential energy = -0.980 J

(c) The spring constant can be calculated as follows:

$-\frac{1}{2}kx^2=-0.980\\ k = \frac{2\times 0.980}{0.08^2} = 306.25 J/m^2$