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## Homework Statement

(a)A spring is attached to a ceiling, and has a relaxed length of 25 cm. When a mass m = 0.80 kg is attached to the spring it stretches to an equilibrium length of L0 = 34 cm.

(b) I lift the mass until the spring returns to its relaxed length, and then release it. When the mass returns to the equilibrium length, what is its speed?

(c) After I release the mass and it falls, what is the length of the spring when the mass reaches its lowest point?

## Homework Equations

F=k*deltay

.5mv^2=kinetic energy

mgy=potential energy

## The Attempt at a Solution

.80*9.81=K*(.34m-.25m)

so K=87.2 for part A

Sorry for the missleading title but i realized im not sure how to do part b or c, i believe for part b i have to set mgy+.5mv^2=mgy+.5mv^2 but im not entirely sure

Want to make sure im doing this right before my quiz tommorow, thanks.

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