Hooke's Law and Work; easy problem, I shouldn't have trouble, but I get it wrong

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SUMMARY

The discussion centers on calculating the work done in stretching a spring according to Hooke's Law. A string with a natural length of 10 inches stretches 1.5 inches under an 8-pound weight, leading to the determination of the spring constant \( k \) as \( \frac{16}{3} \). The work done is computed using the integral \( W = \frac{16}{3} \int_0^4 x \, dx \), resulting in an incorrect value of \( \frac{2048}{9} \) ft-lbs, while the correct answer is \( \frac{128}{3} \) ft-lbs. The error is attributed to arithmetic mistakes during calculations.

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Homework Statement


A string of natural length 10 in. stretches 1.5 in. under a weight of 8-lb. Find the work done in stretching the spring from its natural length to a length of 14 in.

Homework Equations


\begin{equation}
W= \int_a^b \, f(x)dx
\end{equation}
\begin{equation}
f(x)=kx
\end{equation}

The Attempt at a Solution


\begin{equation}
8=k(\frac{3}{2})
\end{equation}
\begin{equation}
k=\frac{16}{3}
\end{equation}
\begin{equation}
W=\frac{16}{3} \int_0^4 \, xdx
\end{equation}
\begin{equation}
W= \frac{16}{3} \, (\frac{1}{2} x^{2} |^{4}_{0})
\end{equation}
\begin{equation}
\frac{2048}{9}
\end{equation}
Book says the answer is:
\begin{equation}
\frac{128}{3} ft-lbs.
\end{equation}
 
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*facepalm* Gotta love arithmetic mistakes.
 

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