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Work done by a spring 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 m from its equilibrium position?

Short Answer

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

Step by step solution

01

Calculate the spring constant k

We are given that a force of \(50 \mathrm{N}\) is required to stretch the spring \(0.5 \mathrm{m}\) from its equilibrium position. We can use Hooke's Law (F = kx) to find the spring constant k: \(k=\frac{F}{x}=\frac{50 \mathrm{N}}{0.5 \mathrm{m}}=100 \mathrm{N/m}\).
02

Calculate the work done when stretching the spring 1.5 m

Using the formula for work done on a spring (W = (1/2)kx^2) and the spring constant k we found in Step 1, we can find the work done when stretching the spring 1.5 m: \(W_{stretch}=\frac{1}{2} \cdot 100 \mathrm{N/m} \cdot (1.5 \mathrm{m})^2=\frac{1}{2} \cdot 100 \mathrm{N/m} \cdot 2.25 \mathrm{m^2}=112.5 \mathrm{J}\).
03

Calculate the work done when compressing the spring 0.5 m

We can use the same formula for work done on a spring (W = (1/2)kx^2) and the spring constant k we found in Step 1, to find the work done when compressing the spring 0.5 m: \(W_{compress}=\frac{1}{2} \cdot 100 \mathrm{N/m} \cdot (0.5 \mathrm{m})^2=\frac{1}{2} \cdot 100 \mathrm{N/m} \cdot 0.25 \mathrm{m^2}=12.5 \mathrm{J}\). So, a. the work done in stretching the spring \(1.5 \mathrm{m}\) from its equilibrium position is \(112.5 \mathrm{J}\), and b. the work done in compressing the spring \(0.5 \mathrm{m}\) from its equilibrium position is \(12.5 \mathrm{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hooke's Law
Underlying many spring-related exercises, including the calculation of work done by a spring, is Hooke's Law. This fundamental principle of physics states that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance. Mathematically, Hooke's Law is expressed as:
\[ F = kx \]
where \( F \) is the force applied to the spring, \( k \) is the spring constant, which measures the stiffness of the spring, and \( x \) is the displacement of the spring from its equilibrium position. When a spring is either compressed or stretched, the force exerted by the spring is directly proportional to its displacement from the equilibrium.
In practical terms, this means that if you double the distance you stretch or compress a spring, the force required to do so also doubles. This linear relationship makes calculations involving springs predictable, which is crucial for applications in engineering and physics.
Spring Constant
The spring constant, often denoted by \( k \), is a measure of the stiffness of a specific spring. It represents the amount of force required to displace the spring by a unit of length. In Hooke's Law (
\[ F = kx \]
), the spring constant \( k \) is the ratio of the force applied on the spring to the displacement produced. The SI unit for the spring constant is newtons per meter (\( N/m \)).
Understanding the spring constant is essential for calculating the work done by a spring, as it factors into the amount of energy stored or released during the spring's displacement. A higher spring constant indicates a stiffer spring that requires more force to stretch or compress. In our exercise example, the spring constant was found by rearranging Hooke's Law to solve for \( k \), giving us a value of \( 100 \, N/m \), which tells us how resistant the spring is to deformation.
Work-Energy Principle
The work-energy principle is a key concept in mechanics that relates the work done on an object to the change in its energy. When applied to springs, the work done in stretching or compressing the spring results in the spring storing potential energy. The formula for the work done on a spring is derived from this principle and is given by:
\[ W = \frac{1}{2}kx^2 \]
This equation describes the amount of work (\( W \)) done to stretch or compress a spring up to a certain distance (\( x \)) from its equilibrium position. The factor \( \frac{1}{2} \) appears because the force exerted by the spring increases linearly from zero (at the equilibrium position) to the maximum force at the full stretch or compression. The energy stored in a spring when displaced is known as elastic potential energy, corresponding to the work required to produce the displacement. Notably, the work done is independent of the process taken to stretch or compress the spring and relies solely on the initial and final positions.
Equilibrium Position
The equilibrium position of a spring is the length at which the spring is neither stretched nor compressed, and thus the force it exerts is zero. It represents the state of the spring where it is at rest and no external forces are acting upon it. When describing the motion of a spring or the work involved in displacing it, the equilibrium position is the reference point from which all measurements of displacement (\( x \)) are taken.
In physics problems, like the calculation of work in springs, recognizing the equilibrium position of the spring serves as the starting point. Displacements are measured relative to this position, whether the spring is compressed or stretched away from equilibrium. Consequently, it's vital to note that whether the spring is stretched or compressed, the work done is always taken in positive values, since work is a scalar quantity and doesn't have a direction associated with it. In the context of the exercise, the 0.5 meters mentioned is the displacement from the equilibrium position, which is crucial for determining the work done on the spring.

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