- #1

Cpt. DeMorgan

- 4

- 0

## Homework Statement

A spring has a relaxed length of 0.17 m, and its stiffness is 15 N/m. The spring sits vertically on a table. You place a block of mass 0.058 kg on the spring and push down on the block until the spring is only 0.11 m long. You hold the block motionless on the compressed spring, then remove your hand.

At the instant you release the block, what is the force exerted by the spring on the block? (Consider the initial stretch of the spring.) Use the usual axis system, with the +y axis running vertically up.

L=0.11m

L0=0.17m

Ks=15N/m

## Homework Equations

|s|(absolute value of s)=|L-L0|

|Fspring|(magnitude of the force) = Ks*|s|

## The Attempt at a Solution

I am asked to find the force exerted by the spring initially. I plugged in the variables I was given.

|s| = |0.11m-0.17m|

|s| = 0.06m

|Fspring| = 15N/m*0.06m

**|Fspring| = 0.9N**

So my answer for the force acted on the block initially by the spring was 0.9N. However, I am told this is not right. Can anyone explain why not?