Question

A spring on a horizontal surface can be stretched and held \(0.5 \mathrm{m}\) from its equilibrium position with a force of \(50 \mathrm{N}\). a. How much work is done in stretching the spring \(1.5 \mathrm{m}\) from its equilibrium position? b. How much work is done in compressing the spring \(0.5 \mathrm{m}\) from its equilibrium position?

Short answer

Question: Calculate the work done in stretching a spring 1.5 m from its equilibrium position (part a) and the work done in compressing the spring 0.5 m from its equilibrium position (part b), given a force of 50 N is required to stretch the spring 0.5 m. Answer: The work done in stretching the spring 1.5 m from its equilibrium position is 112.5 J, and the work done in compressing the spring 0.5 m from its equilibrium position is 12.5 J.

Step-by-step solution

Write down Hooke's Law and the formula for Work done on a spring

Hooke's Law: \(F = kx\) Work done on a spring: \(W = \frac{1}{2}kx^2\)

Calculate the spring constant (k) using the given force and displacement

The force required to stretch the spring 0.5 m is 50 N. We can use Hooke's Law (\(F = kx\)) to calculate the spring constant (k): \(50 = k \times 0.5\) k = \(100\,\text{N/m}\)

Calculate the work done in stretching the spring 1.5 m from its equilibrium position (part a)

For this part of the question, we need to find the amount of work done to stretch the spring 1.5 m from its equilibrium position. Use the formula for work done on a spring: \(W = \frac{1}{2}kx^2\) Plug in the values for k and x (1.5 m): \(W = \frac{1}{2}(100)(1.5)^2\) \(W = \frac{1}{2}(100)(2.25)\) \(W = 112.5 \,\text{J}\) The work done in stretching the spring 1.5 m from its equilibrium position is 112.5 J.

Calculate the work done in compressing the spring 0.5 m from its equilibrium position (part b)

For this part of the question, we need to find the amount of work done to compress the spring 0.5 m from its equilibrium position. Since the spring is symmetrical, the work done to compress it 0.5 m should be the same as the work done to stretch it 0.5 m. Thus, the work done in compressing the spring 0.5 m from its equilibrium position is the same as the work done in stretching it 0.5 m: \(W = \frac{1}{2}(100)(0.5)^2\) \(W = \frac{1}{2}(100)(0.25)\) \(W = 12.5 \,\text{J}\) The work done in compressing the spring 0.5 m from its equilibrium position is 12.5 J.

Key concepts

Spring Constant

A spring constant, denoted as "k," is a measure of how stiff a spring is. It tells us how much force we need to either stretch or compress a spring by a unit distance. It is a crucial part of Hooke's Law, which is represented by the formula: \( F = kx \).- \( F \) stands for the force applied to the spring.- \( k \) is the spring constant.- \( x \) is the displacement from the equilibrium position.When you have a larger spring constant, the spring is stiffer and requires more force to stretch or compress. In our example, a force of 50 N stretches the spring by 0.5 meters.
The equation can be rearranged to find the spring constant:\[ k = \frac{F}{x} \].After substituting the known values (50 N and 0.5 m), we get:\( k = 100 \,\text{N/m} \).Understanding the spring constant helps us predict how a spring will react under different forces, making it a vital concept in mechanics.

Work Done on a Spring

The work done on a spring refers to the energy required to either compress or stretch the spring from its equilibrium position. One can calculate it using the formula:\( W = \frac{1}{2}kx^2 \).This formula calculates the work based on the spring constant \( k \) and the displacement \( x \) from equilibrium:- The factor "\( \frac{1}{2} \)" accounts for the energy stored in the spring as potential energy.- "\( k \)" is the spring constant.- "\( x \)" is the displacement of the spring from its equilibrium point.In our exercise, the work done when stretching or compressing the spring involves plugging values into this formula. For instance, stretching the spring by 1.5 meters yields:\( W = \frac{1}{2}(100)(1.5)^2 = 112.5 \,\text{J} \).Even when compressing it by 0.5 meters, the same calculation applies:\( W = \frac{1}{2}(100)(0.5)^2 = 12.5 \,\text{J} \).Understanding the work done helps us grasp how much energy systems require to maintain or alter spring loads.

Equilibrium Position

The equilibrium position of a spring is the point where it is neither compressed nor stretched. Essentially, it's where the spring is in its neutral state, experiencing no external forces. We often use this position as a reference for measuring displacement. In physics, this concept is important because:

• It serves as the baseline for understanding how forces act upon the spring.
• It allows us to calculate the displacement more accurately.

When a spring is at equilibrium, any task involving stretching or compressing is measured from this neutral point. If you stretch a spring, you pull it away from this equilibrium, and if you compress it, you push it towards and past the equilibrium position.
Knowing this concept allows one to precisely determine how much force is necessary to achieve a specific displacement, which is essential in various applications, like systems that require precise control of movement.