Problem 6 - Thomas Calculus

Question

Bathroom scale A bathroom scale is compressed 1\(/ 16\) in. when a \(150-\) lb person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's Law, how much does someone who compresses the scale 1\(/ 8\) in. weigh? How much work is done compressing the scale 1\(/ 8\) in.?

Step-by-Step Solution

Answer

The person weighs 300 lb, and 18.75 inch-lbs of work is done compressing the scale.

Step 1: Understanding Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its compression distance. It can be expressed as \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the compression distance. We need to solve for \( k \), the spring constant, using the initial compression scenario.

Step 2: Calculate the Spring Constant

Given that a 150 lb person compresses the scale 1/16 inch, we use Hooke's Law: \( 150 = k \times \frac{1}{16} \). Solving for \( k \), we multiply both sides by 16, yielding \( k = 2400 \) lb/in.

Step 3: Determine Weight for New Compression

To find the weight of a person compressing the scale by 1/8 inch, again use Hooke's Law: \( F = kx \). Substitute \( k = 2400 \) lb/in and \( x = 1/8 \) inch into the equation, resulting in \( F = 2400 \times \frac{1}{8} = 300 \) lb.

Step 4: Calculate Work Done on Spring

Work done on a spring is calculated using the formula \( W = \frac{1}{2}kx^2 \). We know \( k = 2400 \) lb/in and \( x = 1/8 \) inch. Substitute these values: \( W = \frac{1}{2} \times 2400 \times \left(\frac{1}{8}\right)^2 \). Calculating this yields \( W = 18.75 \) inch-lbs.

Key Concepts

Spring Constant

Hooke's Law helps us understand the relationship between force and compression in springs. The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is required to compress or stretch the spring by a certain distance. To find the spring constant, we use the formula from Hooke's Law: \( F = kx \), where:

• \( F \) is the force exerted on the spring
• \( k \) is the spring constant
• \( x \) is the compression distance

In the given problem, a 150 lb person compresses the scale by \( \frac{1}{16} \) inch. By rearranging the formula to solve for \( k \), we get:\[ k = \frac{F}{x} \]Substituting the known values gives us \( k = \frac{150}{\frac{1}{16}} = 2400 \) lb/in. This means the spring requires 2400 pounds of force to be compressed by one inch.

Work Done

The work done in compressing a spring is an important concept when considering how much energy is used in the process. Work done by a force on a spring involves deforming it, either through compression or elongation. The formula for calculating this work is:\[ W = \frac{1}{2}kx^2 \]Where:

• \( W \) is the work done
• \( k \) is the spring constant
• \( x \) is the compression or extension distance from the spring's equilibrium position

In our exercise, we already determined that the spring constant \( k \) is 2400 lb/in. When the scale is compressed by \( \frac{1}{8} \) inch, the work done is:\[ W = \frac{1}{2} \times 2400 \times \left(\frac{1}{8}\right)^2 \]Calculating this, we find \( W = 18.75 \) inch-lbs. This value represents the energy required to compress the spring by \( \frac{1}{8} \) inch.

Compression Distance

Compression distance is a key factor in Hooke's Law as it determines how much a spring is compressed under a given force. The distance a spring is compressed or stretched refers to how far it is moved from its original, unloaded position.In the exercise, the compression distances mentioned are \( \frac{1}{16} \) inch and \( \frac{1}{8} \) inch. It's crucial to note that although these measurements seem small, even minute changes in compression distance can significantly affect the force applied and the work done on the spring.For example, when the scale was compressed by \( \frac{1}{16} \) inch, a force of 150 lb was applied. When compressed by \( \frac{1}{8} \) inch, a greater force of 300 lb was necessary due to the linear nature of Hooke's Law. Understanding the compression distance helps us predict how the spring behaves under different loads, enabling accurate calculations for both mechanical and structural applications.