# Spring question(just need answer checked)

## 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.

Last edited:

rl.bhat
Homework Helper
Your calculation of part A is correct.
Consider the equilibrium position as the reference point for measuring PE with zero initial PE.
When you lift the mass through a distance x, the rise in PE = mgx.
When you release the mass its initial velocity is zero.
When it crosses the equilibrium position, fall in PE = mgx
This change in PE in converted in to KE of mass and elastic potential energy in spring.
Write down the equation and solve for v.
For the last part, the final velocity of the mass is zero.
So change in KE = fall in PE + increase in the elastic PE
Solve this equation to find the final extension of the mass..