Calculating work using hookes law

[dw10]

Homework Statement

Calculate the work done, in Joules for a system in which a muscle of 1cm^2 cross section and 10cm length is stretched to 11cm by hanging a mass on it. The muscle behaves like a spring. The spring constant for the muscle was determined by finding that the muscle exerts a force of 5.00N when it is stretched from 10.0cm to 10.5cm

Homework Equations

Is spring constant (-k) calculated simply by dividing .005m into 5.00N? (because F= -kx, and -k = F/x) = -1000.

The Attempt at a Solution

i don't think my answer will be correct because i have a feeling that I am calculating k incorrectly??
OK so assuming I am doing it right though...

i set up the integral using hookes law:

|work| = int (from .10m to .11m) -k (x - x0)dx

= int (from .10m to .11m) -1000 (x - x0)dx
then i took constants out of integral and integrated...

|work| = -1000/2 * |from .10m to .11m [ (xf-x0)^2 - (xi - x0)^2]

i don't want to proceed further unless i am on right track (which i doubt) thanks for any help!

 

[Dick]

Right. k=1000*kg/s^2. Put units on the these numbers, ok? They aren't dimensionless. Now continue with the second part.

 

[kerimek]

I would first clarify the units being used in this problem. The spring constant, k, should have units of N/m, not N/cm as suggested in the attempted solution.

To calculate the work done by the muscle, we can use the formula W = 1/2 * k * (x2^2 - x1^2), where k is the spring constant and x2 and x1 are the final and initial lengths of the muscle, respectively.

In this case, x2 = 11cm and x1 = 10cm, so the work done would be:

W = 1/2 * (5 N/cm) * [(11 cm)^2 - (10 cm)^2] = 5.5 J

It is important to use consistent units in calculations to ensure the accuracy of the result.

Additionally, the spring constant should not be calculated by dividing 5.00N by 0.005m, as the units do not match (N/m vs. N/cm). The correct way to determine the spring constant would be to rearrange the equation F = -kx and solve for k using the given values of force and displacement.