Problem 5 - Basic Technical Mathematics with Calculus

Question

A 160 -lb person compresses a bathroom scale 0.080 in. If the scale obeys Hooke's law, how much work is done compressing the scale if a 180 -lb person stands on it?

Step-by-Step Solution

Answer

The work done compressing the scale is 8.1 in-lb.

Step 1: Understand Hooke's Law

Hooke's Law states that the force required to compress or extend a spring is directly proportional to the distance it is compressed or extended. Mathematically, it's defined as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the distance compressed or extended.

Step 2: Determine the Spring Constant

First, use Hooke's Law with the data given for the 160-lb person to find the spring constant \( k \). The force is 160 lb and the distance is 0.080 in. Thus, \( 160 = k \times 0.080 \). Calculate \( k \) by \( k = \frac{160}{0.080} \).

Step 3: Calculate the Spring Constant

Perform the division to find \( k \): \( k = \frac{160}{0.080} = 2000 \) lb/in.

Step 4: Use Hooke's Law for 180-lb Person

Apply Hooke’s Law again to find the compression when a 180-lb person stands on the scale: \( 180 = 2000 \times x \). Solve for \( x \) by \( x = \frac{180}{2000} \).

Step 5: Calculate the New Compression Distance

Divide to find \( x \): \( x = \frac{180}{2000} = 0.090 \) in.

Step 6: Determine Work Done on the Scale

The work done on a spring is given by the formula \( W = \frac{1}{2} k x^2 \). Use the spring constant \( k = 2000 \) lb/in and distance \( x = 0.090 \) in.

Step 7: Calculation of Work Done

Substitute the values into the work formula: \( W = \frac{1}{2} \times 2000 \times (0.090)^2 \). First, calculate \( (0.090)^2 = 0.0081 \), then \( W = 0.5 \times 2000 \times 0.0081 = 8.1 \) in-lb.

Key Concepts

Spring Constant

The spring constant, denoted as \( k \), plays a vital role in Hooke's Law. It quantifies the stiffness of a spring. A higher spring constant indicates a stiffer spring, requiring more force to compress it the same distance compared to a spring with a lower constant. When working with Hooke's Law, the formula \( F = kx \) helps determine how much force is needed to compress or extend a spring. Here, \( F \) is the force applied, and \( x \) is the distance moved. For the bathroom scale problem, the spring constant was calculated using the weight of a person and the compression caused. By rearranging the formula, we find \( k \) through \( k = \frac{F}{x} \). Understanding this concept is crucial in ensuring accurate calculations in problems involving springs and other elastic materials.

Work Done

The concept of work done refers to the energy transferred when an object is moved over a distance by a force. In the context of springs, this is described by the equation \( W = \frac{1}{2} k x^2 \). This formula calculates the energy required to compress or extend a spring a certain distance. When we talk about the bathroom scale, work done is the energy used to compress the scale. The formula \( W = \frac{1}{2} k x^2 \) incorporates the spring constant and the distance, \( x \), which is the amount the spring is compressed. Work done is an essential aspect of understanding how potential energy is stored in springs. It explains why we sometimes need significant force to compress stiff springs even a little.

Force and Compression

Force and compression are central to understanding how springs work with Hooke's Law. Force, measured in pounds or newtons, refers to any interaction that, when unopposed, will change the motion of an object. Compression, in this scenario, is the action of reducing the length of a spring due to an applied force.Hooke's Law succinctly states this relationship through the equation \( F = kx \). As the force increases, the compression increases, assuming the spring constant is constant. This linear pattern is helpful in predicting and analyzing physical interactions across various practical situations. In the given exercise, the compression was determined by using the formula, demonstrating the direct relationship between the weight applied and the distance the scale compresses.

Application of Physics in Mathematics

Physics and mathematics often intertwine, especially when we analyze forces and motions mathematically. This exercise exemplifies how mathematical formulas model physical concepts. Using Hooke's Law and the work done formula, students can break down the behavior of objects under force into calculable formats. This not only solves practical problems, such as the compression of a scale under different weights, but it also provides deeper insights into the connection between theoretical equations and real-world scenarios. By understanding how formulas like \( F = kx \) and \( W = \frac{1}{2} k x^2 \) are derived and applied, students enhance their ability to translate physical observations into mathematical processes, enriching both their physics and mathematics education.