Question Statement

It takes a force of $50\text{ N}$ to stretch a spring $0.5\text{ m}$. Find the work done in stretching the spring $0.6\text{ m}$ beyond its natural length.

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Background and Explanation

This problem applies Hooke's Law, which states that the force required to stretch or compress a spring is proportional to the displacement from its natural length. Since the force varies with position, we must use integration to calculate the work done.

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Solution

Step 1: Determine the Spring Constant

According to Hooke's Law, the force $F$ required to stretch a spring by distance $x$ from its natural length is:

$F=kx$

where $k$ is the spring constant.

Given that a force of $50\text{ N}$ produces an extension of $0.5\text{ m}$:

$50=k(0.5)$

Solving for $k$:

$k=\frac{50}{0.5}=100\text{ N/m}$

Therefore, the force function is:

$F(x)=100x$

Step 2: Set Up the Work Integral

The work done by a variable force $F(x)$ in moving an object from position $x=a$ to $x=b$ is given by:

$W={\int }_a^{b}F(x)dx$

For a spring stretched from its natural length ($x=0$) to $x=0.6\text{ m}$:

$W={\int }_0^{0.6}100xdx$

Step 3: Evaluate the Integral

Compute the definite integral:

\begin{align*} W &= 100 \int_{0}^{0.6} x \, dx \\ &= 100 \left[ \frac{x^2}{2} \right]_{0}^{0.6} \\ &= \frac{100}{2} \left[ x^2 \right]_{0}^{0.6} \\ &= 50 \left( (0.6)^2 - 0^2 \right) \\ &= 50(0.36) \\ &= 18 \text{ J} \end{align*}

Final Answer

The work done in stretching the spring $0.6\text{ m}$ beyond its natural length is $\boxed{18\text{ J}}$.

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Key Formulas or Methods Used

• Hooke's Law: $F=kx$ (Force is proportional to displacement from natural length)
• Spring Constant Calculation: $k=\frac{F}{x}$
• Work by Variable Force: $W={\int }_{x_1}^{x_2}F(x)dx$
• Power Rule for Integration: $\int xdx=\frac{x^2}{2}+C$

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Summary of Steps

1. Find $k$: Use the given force and extension to calculate the spring constant: $k=\frac{50}{0.5}=100\text{ N/m}$
2. Define $F(x)$: Write the force function: $F(x)=100x$
3. Set up integral: Write the work formula with proper limits: $W={\int }_0^{0.6}100xdx$
4. Integrate: Evaluate to get $W=50[x^2{]}_0^{0.6}$
5. Calculate: Substitute bounds: $W=50(0.36)=18\text{ J}$