Solving the Spring Balance Problem: Work Done Stretching 5.0-12.2in

[thehsheepdog]

Hi all,

I could use some assistance with the following problem.

"The spring of a spring balance is 5.0in. long when there is no weight on the balance, and it is 9.6in. long with 8.0 lb hung from the balance. How much work is done stretching it from 5.0in to 12.2in?"

So far, I know to take the constant by solving for K:

\(\displaystyle 9.6=8.0k\) which \(\displaystyle =1.2\)

Then, I know that this problem needs to be integrated from 5.0 to 12.2 inches

\(\displaystyle \int_5^{12.2}\)

And this is where I am stumped. I don't know if I am integrating the constant (K) or something else?Thank you for your help.

 

[topsquark]

Hi all,

I could use some assistance with the following problem.

"The spring of a spring balance is 5.0in. long when there is no weight on the balance, and it is 9.6in. long with 8.0 lb hung from the balance. How much work is done stretching it from 5.0in to 12.2in?"

So far, I know to take the constant by solving for K:

\(\displaystyle 9.6=8.0k\) which \(\displaystyle =1.2\)

Then, I know that this problem needs to be integrated from 5.0 to 12.2 inches

\(\displaystyle \int_5^{12.2}\)

And this is where I am stumped. I don't know if I am integrating the constant (K) or something else?Thank you for your help.

Work done by a variable force is given by
\(\displaystyle \int \textbf{F} \cdot d \textbf{s}\)

In this case \(\displaystyle \textbf{F} = -k \textbf{x}\), where the positive direction is defined in the direction of the compression of the spring. Since you are stretching the spring F is in the direction of the compression and x is in the direction of the stretch, \(\displaystyle \textbf{F} \cdot d \textbf{s}\) is always positive. So:
\(\displaystyle W = \int_{5}^{12.2} -k \textbf{x} \cdot d \textbf{x} = k \int_5^{12.2}x~dx\)

\(\displaystyle W = \left . \frac{1}{2}kx^2 \middle | _{5}^{12.2} \right .\)

-Dan