Bear asked in Science & MathematicsPhysics · 1 month ago

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. 

1 Answer

  • Ash
    Lv 7
    1 month ago
    Favorite 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

Still have questions? Get your answers by asking now.