Question Statement

A force $F=\frac{3}{2}x$ lb is needed to stretch a 10-inch spring an additional $x$ inches. Find the work done in stretching the spring 16 inches.

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Background and Explanation

This problem applies the concept of work done by a variable force, which requires integration rather than simple multiplication. When force varies with displacement (as described by Hooke's Law-type relationships), we calculate work by integrating the force function over the interval of stretch.

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Solution

Interpreting the Problem

The force function is given as $F(x)=\frac{3}{2}x$ lb, where $x$ represents the additional stretch in inches beyond the spring's natural length. To find the work done in stretching the spring by 16 inches (interpreted as an additional 16 inches of stretch), we integrate from $x=0$ to $x=16$.

Setting Up the Integral

The work $W$ done by a variable force is defined as: $W={\int }_a^{b}F(x)dx$

Substituting the given force function and limits: $\text{Work done}={\int }_0^{16}\frac{3}{2}xdx$

Evaluating the Integral

Factor out the constant $\frac{3}{2}$: $=\frac{3}{2}{\int }_0^{16}xdx$

Apply the power rule for integration $\int xdx=\frac{x^2}{2}$: $=\frac{3}{2}{\left[\frac{x^2}{2}\right]}_0^{16}$

Simplify the constant coefficient: $=\frac{3}{4}{\left[x^2\right]}_0^{16}$

Calculating the Definite Integral

Evaluate at the upper and lower limits: $=\frac{3}{4}\left((16{)}^2-(0{)}^2\right)$ $=\frac{3}{4}(256-0)$ $=\frac{3}{4}(256)$

Perform the final multiplication: $=3\times 64=192\text{ lb}$

Therefore, the work done in stretching the spring 16 inches is 192 inch-pounds (or 192 lb).

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Key Formulas or Methods Used

• Work by a Variable Force: $W={\int }_a^{b}F(x)dx$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Definite Integral Evaluation: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is the antiderivative of $f$

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Summary of Steps

1. Identify the force function $F(x)=\frac{3}{2}x$ and determine the integration limits from $x=0$ to $x=16$ inches (representing the additional stretch).
2. Set up the work integral: $W={\int }_0^{16}\frac{3}{2}xdx$.
3. Factor out constants: $\frac{3}{2}{\int }_0^{16}xdx$.
4. Find the antiderivative: $\frac{3}{2}\cdot \frac{x^2}{2}=\frac{3}{4}{x}^2$.
5. Evaluate the definite integral: $\frac{3}{4}(16^2-0^2)=\frac{3}{4}(256)$.
6. Compute the final value: $3\times 64=192$ lb.