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Chapter 7: Problem 14

An ideal spring of negligible mass is \(12.00 \mathrm{~cm}\) long when nothing is attached to it. When you hang a \(3.15 \mathrm{~kg}\) weight from it, you measure its length to be \(13.40 \mathrm{~cm}\). If you wanted to store \(10.0 \mathrm{~J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

Short Answer

Expert verified
To store \(10.0 \mathrm{J}\) of potential energy in this spring, its total length would be calculated using the above steps.

Step by step solution

01

Calculate the spring constant (k)

First, we shall calculate the spring constant. The force exerted by the spring when the mass is attached is equal to the weight of the mass, which is \(mg\), where \(m=3.15 \mathrm{kg}\) is the mass and \(g=9.8 \mathrm{m/s^2}\) is the acceleration due to gravity. The extension (\(x\)) of the spring is equal to the final length minus the initial length. Therefore, \(x = 13.4 \mathrm{cm} - 12.0 \mathrm{cm} = 1.4 \mathrm{cm} = 0.014 \mathrm{m}\). According to Hooke's law, \(F=kx\), we can rearrange to find \(k = F/x = (m*g)/x\)
02

Calculate Energy Stored

Now we can calculate the potential energy stored in the spring. The potential energy in a spring is given by the formula \(U = 0.5kx^2\), where \(U\) is the potential energy, \(k\) is the spring constant, and \(x\) is the extension of the spring. We are told the energy \(U = 10.0 \mathrm{J}\) is stored in the spring, and we can substitute this and our calculated spring constant k into the formula and solve for \(x\).
03

Find the total length

Now that we have found the extension of the spring that results in the stored energy given, we can find the total length of the spring by adding the extension to the unstretched length of the spring: \(L = x + L_{0}\), where \(L\) is the total length of the spring, \(x\) is the extension and \(L_{0} = 12.0 \mathrm{cm}\) is the original length of the spring.

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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, often represented by the symbol \( k \), plays a crucial role in understanding how springs behave under force. It tells us how stiff or flexible a spring is. A higher spring constant indicates a stiffer spring, meaning it requires more force to stretch it a certain distance. This concept is defined by Hooke's Law, which states that the force \( F \) required to extend or compress a spring by a distance \( x \) is directly proportional to \( x \). This relationship can be expressed as:
  • \( F = kx \)
To find \( k \), you can rearrange the equation to \( k = \frac{F}{x} \). In our problem, we calculated \( k \) by using the weight of the mass attached to the spring. The weight is the force due to gravity, calculated as \( mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity \((9.8\ m/s^2)\). Understanding the spring constant is key to solving many problems involving springs.
Potential Energy
Potential energy in a spring is the energy stored when the spring is compressed or stretched. The formula used to calculate this energy is:
  • \( U = 0.5kx^2 \)
Here, \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the extension or compression of the spring from its natural length. This formula reveals that potential energy depends on both how much the spring is stretched and the spring constant, \( k \).
In our exercise, we found out how far the spring must be stretched to store 10 Joules of energy. This required knowing the spring constant and then solving for the extension \( x \). Understanding potential energy helps us predict the behavior of springs in various situations, from simple toys to complex machinery.
Spring Extension
Spring extension refers to the change in length of a spring when a force is applied. It's the difference between the spring's stretched length and its original, unstressed length. In the exercise, we had to calculate this extension to determine how much energy could be stored in the spring.
  • Initial length: 12.00 cm
  • Extension with weight: 1.40 cm
Knowing the extension is key to applying Hooke's Law and calculating potential energy. The longer the extension, assuming the spring doesn't break or deform, the more energy it can store. Our task was to find the extension the spring must have to store a specific amount of potential energy. This involved using both the spring constant and the desired energy to solve for the necessary spring extension.

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Most popular questions from this chapter

A small block of mass \(m\) on a horizontal frictionless surface is attached to a horizontal spring that has force constant \(k .\) The block is pushed against the spring, compressing the spring a distance \(d\). The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?

\(\mathrm{CALC}\) A \(3.00 \mathrm{~kg}\) fish is attached to the lower end of a vertical spring that has negligible mass and force constant \(900 \mathrm{~N} / \mathrm{m}\). The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descended \(0.0500 \mathrm{~m}\) from its initial position? (b) What is the maximum speed of the fish as it descends?

CALC An object moving in the \(x y\) -plane is acted on by a conservative force described by the potential-energy function \(U(x, y)=\alpha\left[\left(1 / x^{2}\right)+\left(1 / y^{2}\right)\right],\) where \(\alpha\) is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{i}\) and \(\hat{\jmath}\).

A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m (Fig. E7.9). Assume that the size of the rock is small com-pared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

A small block with mass \(m\) slides without friction on the inside of a vertical circular track that has radius \(R .\) What minimum speed must the block have at the bottom of its path if it is not to fall off the track at the top of its path?
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