Problem 15 - University Physics with Modern Physics

Question

A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m} .\) (a) What is the potential energy of the spring when it is stretched 0.200 \(\mathrm{m} ?(\mathrm{b})\) What is its potential energy when it is compressed 5.00 \(\mathrm{cm} ?\)

Step-by-Step Solution

Answer

(a) 80 J when stretched by 0.200 m; (b) 5 J when compressed by 5.00 cm.

Step 1: Understand the Problem

We have a spring with a known force and displacement. We need to calculate the spring's potential energy when it is stretched and when it is compressed. The spring constant can be found using Hooke's Law, and the potential energy can be calculated using the formula for elastic potential energy.

Step 2: Find the Spring Constant

Use Hooke's Law, which states \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement. Rearrange to find \( k = \frac{F}{x} \). Substitute \( F = 800 \) N and \( x = 0.200 \) m to get \( k = \frac{800}{0.200} = 4000 \) N/m.

Step 3: Calculate Potential Energy When Stretched

Use the formula for elastic potential energy: \( PE = \frac{1}{2} kx^2 \). Substitute \( k = 4000 \) N/m and \( x = 0.200 \) m to find \( PE = \frac{1}{2} \times 4000 \times (0.200)^2 = 80 \) J.

Step 4: Convert Compression Distance and Calculate Potential Energy When Compressed

Convert the compression distance from 5.00 cm to meters: \( 0.050 \) m. Use \( PE = \frac{1}{2} kx^2 \) again. Substitute \( k = 4000 \) N/m and \( x = 0.050 \) m to find \( PE = \frac{1}{2} \times 4000 \times (0.050)^2 = 5 \) J.

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a fundamental concept derived from Hooke's Law. It represents the stiffness of a spring, indicating how much force is needed to stretch or compress the spring by a certain distance. In the exercise, Hooke's Law is used to calculate the spring constant with the formula \( F = kx \). Here, \( F \) represents the force applied, \( k \) is the spring constant, and \( x \) stands for the displacement. Rearranging this formula gives \( k = \frac{F}{x} \),which tells us how resistant a spring is to being deformed.

This resistance is consistent across the entire stretch of the spring, so a high spring constant means you need more force to achieve the same amount of stretch compared to a spring with a lower constant. Understanding the spring constant helps in predicting how a spring will behave under different forces.

• Hooke's Law: \( F = kx \)
• Spring Constant Formula: \( k = \frac{F}{x} \)
• Measures the stiffness of a spring

Elastic Potential Energy

Elastic potential energy refers to the energy stored in a spring when it is either compressed or stretched. This energy is given back when the spring returns to its original shape. In the exercise, the formula used to calculate elastic potential energy is \( PE = \frac{1}{2} kx^2 \), where \( PE \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.

The main takeaway is that the energy depends on both the square of the displacement and the spring constant. This means if you double the displacement, the energy stored increases fourfold, because it is proportional to the square of that displacement. Elastic potential energy is essential for understanding how springs and other elastic materials behave in mechanical systems.

• Formula: \( PE = \frac{1}{2} kx^2 \)
• Energy depends on displacement squared
• Energy stored during deformation

Elasticity

Elasticity refers to the ability of a material to return to its original shape after being stretched or compressed. It is a crucial aspect of many materials and objects, such as springs, rubber bands, and even human tissues. In our exercise context, elasticity explains why the spring returns to its original state after the force is removed.

Materials with high elasticity, like a slinky or a high-quality spring, can undergo significant changes in shape without permanent deformation. This property is not infinite; a material will eventually reach a point where it can no longer return to its original shape and may permanently deform or even break. Understanding elasticity is crucial when calculating how materials behave under force and is essential in determining suitable applications for different materials.

• Ability to return to original shape
• Important for materials like springs
• Limit to elasticity called 'elastic limit'