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# 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

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- 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

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