- #1

8008jsmith

- 15

- 0

## Homework Statement

A muscle of 1 cm^2 cross section and 10 cm length is stretched to 11 cm by hanging a mass on it. The muscle behaves like a spring that follows Hooke's law. The Hooke's law constant for the muscle was determined by finding that the muscle exerts a force of 5 N when it is stretched from 10 cm to 10.5 cm.

## Homework Equations

The formula I used was: w = 1/2k (x2 - x0)^2 - (x1-x0)^2

## The Attempt at a Solution

I came up with -20 Joules for the answer but I'm not sure if this is right. I don't know why they included the cross section.

First I solved for k using the first set of numbers. So... 5 N = 1/2 k (0.105m - 0.1m)^2 - (0.1m - 0.1m)^2

But I'm not sure if that is right because (w) should be in Newton meters not just Newtons, right?