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To stretch a spring 3.00 \(\mathrm{cm}\) from its unstretched length, 12.0 \(\mathrm{J}\) of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 \(\mathrm{cm}\) from its unstretched length? (c) How much work must be done to compress this spring 4.00 \(\mathrm{cm}\) from its unstretched length, and what force is needed to streteh it this distance?

Short Answer

Expert verified
(a) Spring constant \( k = 26666.67 \, \text{N/m} \). (b) Force needed is 800 N. (c) Work is 21.33 J; force required is 1066.67 N.

Step by step solution

01

Understanding Hooke's Law

Hooke's Law states the force required to stretch or compress a spring by distance \( x \) is proportional to that distance. Mathematically, it's given by \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the change in length from the equilibrium position.
02

Determine Spring Constant

To find the spring constant \( k \), use the work done on the spring. The work done (\( W \)) to stretch or compress a spring is given by \( W = \frac{1}{2}kx^2 \). Rearrange to find \( k \):\[ k = \frac{2W}{x^2}\]Given \( W = 12.0 \) J and \( x = 0.03 \) m (3 cm converted to meters), substitute the values in:\[k = \frac{2 \times 12.0}{(0.03)^2} = 26666.67 \, \text{N/m}\]
03

Calculate Force for 3.00 cm Extension

Using Hooke's Law \( F = kx \) and with \( k = 26666.67 \, \text{N/m} \) and \( x = 0.03 \, \text{m} \):\[ F = 26666.67 \times 0.03 = 800 \, \text{N}\]810 N of force is needed to stretch the spring 3.00 cm.
04

Calculate Work to Compress 4.00 cm

For compression, still use the work formula \( W = \frac{1}{2}kx^2 \). Given \( k = 26666.67 \, \text{N/m} \) and \( x = 0.04 \, \text{m} \):\[ W = \frac{1}{2} \times 26666.67 \times (0.04)^2 = 21.33 \, \text{J}\]21.33 J of work must be done to compress the spring 4.00 cm.
05

Calculate Force for 4.00 cm Compression

Use Hooke's Law \( F = kx \) for \( x = 0.04 \, \text{m} \):\[ F = 26666.67 \times 0.04 = 1066.67 \, \text{N}\]1066.67 N of force is needed to compress the spring 4.00 cm.

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Key Concepts

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

Spring Constant
The spring constant, denoted as \( k \), measures the stiffness of a spring. It is a critical component in understanding how much force is needed to stretch or compress the spring. Imagine pulling a rubber band; if it stretches easily, it has a low spring constant. Conversely, a stiffer, harder-to-pull spring has a higher \( k \). To determine this constant, we need to understand Hooke's Law, expressed as \( F = kx \), where \( F \) stands for the force applied, \( k \) is the spring constant, and \( x \) represents the displacement from the equilibrium position.

When we know the work done to stretch or compress the spring, we can use the work-energy relationship to find \( k \). The equation \( W = \frac{1}{2}kx^2 \) quantifies the work done, where \( W \) is the work in joules. By rearranging this equation, we express the spring constant as \( k = \frac{2W}{x^2} \). Using this formula with the given values allows us to solve real-life problems and understand the intrinsic properties of a spring's behavior.
Work-Energy Principle
The work-energy principle links the work done on an object to its energy changes. This principle is particularly useful in spring mechanics. When a force stretches or compresses a spring, it transfers energy into the spring's potential energy storage, analogous to a battery being charged.

For springs, the energy stored or work done is calculated as \( W = \frac{1}{2}kx^2 \). Here, \( W \) stands for work or energy in joules, \( k \) represents the spring constant, and \( x \) is the displacement. This equation demonstrates that the work required to stretch or compress a spring is directly proportional to the square of the distance the spring is moved.

Understanding this relationship helps explain why doubling the stretch of a spring requires four times the work, illustrating the nonlinear accumulation of energy in spring mechanics. This concept not only aids in solving physics problems but also in designing systems where springs are used, such as automotive suspensions or trampoline manufacturing.
Mechanics
Mechanics, a branch of physics, deals with forces and motion. It provides the framework for understanding the motion of objects and the forces acting on them. Within mechanics, springs are modeled through Hooke's Law, and their behavior is analyzed using fundamental principles.

When dealing with problems involving springs, mechanics allow us to calculate forces and energy changes efficiently. For example, using the spring constant from Hooke's Law \( F = kx \) helps determine the force needed to achieve a specific displacement. Furthermore, mechanics incorporates various formulas, such as those for calculating work or kinetic and potential energy, to assess the total mechanical energy within a system involving springs.

The concepts in mechanics guide us in understanding practical applications, from everyday objects like shock absorbers and door closers to specialized contexts such as seismic isolation systems in earthquake-prone zones. By grasping these principles, one can innovate and create solutions that incorporate elastic potential energy effectively.

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Most popular questions from this chapter

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