constant of the spring?
A 6.5 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.0 cm if the marble is to just reach a target 14 m above the marble's position on the compressed spring. The change in the gravitational potential energy of the marble-Earth system during the 14 m ascent? is .891J.
What is the constant of the spring?
I can't seem to figure this one out.
- AshLv 71 month agoFavorite Answer
Change in PE = mgh = (0.0065 kg)(9.8 m/s²)(14 m) = 0.892 J ( with 3 sig fig)
Spring PE = change in gravitational PE
½kx² = 0.892
k = 2(0.892)/x²
k = 2(0.892)/(0.09)²
k = 220 N/m