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\(\bullet\) To stretch a certain spring by 2.5 \(\mathrm{cm}\) from its equilibrium position requires 8.0 \(\mathrm{J}\) of work. (a) What is the force constant of this spring? (b) What was the maximum force required to stretch it by that distance?

Short Answer

Expert verified
The spring constant is 25,600 N/m and the maximum force is 640 N.

Step by step solution

01

Understand the Relationship Between Work and Spring Constant

When a spring is stretched or compressed, the work done on the spring relates to the spring constant (\( k \)) and the displacement (\( x \)) through the equation for elastic potential energy: \[ W = \frac{1}{2} k x^2 \]. Here, \( W \) is the work done in joules, \( k \) is the spring constant in N/m, and \( x \) is the displacement from the equilibrium position in meters.
02

Convert Units

Before calculating the spring constant, ensure that the displacement \( x \) is in meters. Given that the displacement is 2.5 cm, convert it to meters: \( x = 2.5 \text{ cm} = 0.025 \text{ m} \).
03

Calculate the Spring Constant

Substitute the values into the equation for the work done: \[ 8.0 = \frac{1}{2} k (0.025)^2 \]. Multiply both sides by 2 to get rid of the fraction: \[ 16 = k (0.025)^2 \]. Solve for \( k \): \[ k = \frac{16}{0.025^2} = 25,600 \text{ N/m} \].
04

Understand the Maximum Force Formula

The maximum force \( F \) required to stretch the spring can be found using Hooke's Law: \( F = kx \). Here, \( k \) is the spring constant, and \( x \) is the displacement in meters.
05

Calculate the Maximum Force

Use the spring constant \( k = 25,600 \text{ N/m} \) and the displacement \( x = 0.025 \text{ m} \) to calculate the force: \[ F = 25,600 \times 0.025 = 640 \text{ N} \].

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Key Concepts

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

Elastic Potential Energy
When we talk about elastic potential energy, we're discussing the energy stored in a spring when it's either stretched or compressed. The amount of energy depends on two main factors: the displacement of the spring from its equilibrium position and the spring constant, which tells us how stiff the spring is.
The formula to calculate the elastic potential energy is given by:
  • \[ W = \frac{1}{2} k x^2 \]
In this equation,
  • \( W \) represents the work done on the spring, measured in joules.
  • \( k \) is the spring constant, measured in newtons per meter (N/m).
  • \( x \) is the displacement from the equilibrium position, measured in meters.
The value of \( W \) will tell you how much energy is stored in the spring when it is either stretched or compressed by a distance \( x \). Understanding this concept can be very useful when you want to determine how much energy is needed to stretch or compress a spring a certain amount.
Hooke's Law
Hooke's Law is a fundamental principle that explains how the force needed to either stretch or compress a spring is related to the spring's characteristics.
According to Hooke's Law, the force required to stretch or compress a spring is directly proportional to the displacement of the spring. The formula is simply:
  • \[ F = kx \]
In this formula:
  • \( F \) is the force applied to the spring, measured in newtons (N).
  • \( k \) is the spring constant, or the "stiffness" of the spring, in N/m.
  • \( x \) is the displacement from the spring's equilibrium position in meters.
What Hooke's Law tells us is that a stiffer spring (higher \( k \)) will require more force to be stretched or compressed the same distance compared to a less stiff one. This law is vital for calculating the mechanical behavior of springs in various applications, such as in automotive suspensions or mattress springs.
Work Done on Spring
The concept of work done on a spring involves the energy transferred to the spring to change its shape, whether that means stretching it or compressing it. Work is done when a force moves an object over a distance. In the case of springs, this distance is how far the spring is stretched or compressed.
To determine the work done on a spring, we use the formula for elastic potential energy:
  • \[ W = \frac{1}{2} k x^2 \]
Here, work \( W \) is equivalent to the energy stored in the spring. It shows us how much energy is "pumped" into the spring to stretch or compress it.
When you apply force to a spring, you compress or stretch it from its neutral or equilibrium position. The work done is useful because it helps us understand how much energy is needed to compress or stretch a spring to a particular point, and this energy is recoverable when the spring returns to its equilibrium position.
Understanding how work relates to springs is important for designing systems that use springs effectively, allowing us to predict how much energy a spring can store or release during operation.

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Most popular questions from this chapter

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