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A spring of negligible mass has force constant \(k=1600 \mathrm{~N} / \mathrm{m}\). (a) How far must the spring be compressed so that \(3.20 \mathrm{~J}\) of potential energy is stored in it? (b) You place the spring vertically with one end on the floor. You then drop a \(1.20 \mathrm{~kg}\) book onto it from a height of \(0.80 \mathrm{~m}\) above the top of the spring. Find the maximum distance the spring will be compressed.

Short Answer

Expert verified
(a) The spring must be compressed by 0.0632 m; (b) The spring compresses 0.1085 m when the book is dropped.

Step by step solution

01

Understand Potential Energy in a Spring

The potential energy stored in a compressed or stretched spring is given by the formula: \( PE = \frac{1}{2} k x^2 \), where \(k\) is the spring constant and \(x\) is the compression or extension distance. Given \(PE = 3.20\, \text{J}\) and \(k = 1600\, \text{N/m}\), we need to find \(x\).
02

Solve for Compression Distance

Rearrange the potential energy equation to solve for \(x\):\[ x = \sqrt{\frac{2 \times PE}{k}} \]Substitute \(PE = 3.20\, \text{J}\) and \(k = 1600\, \text{N/m}\) into the equation:\[ x = \sqrt{\frac{2 \times 3.20}{1600}} = \sqrt{\frac{6.4}{1600}} = \sqrt{0.004} = 0.0632\, \text{m} \]Thus, the spring must be compressed by \(0.0632\, \text{m}\) or \(6.32\, \text{cm}\) for \(3.20\, \text{J}\) of energy to be stored.
03

Calculate Energy from Dropping the Book

When the book is dropped, its initial gravitational potential energy is converted into spring potential energy. The potential energy due to gravity is given by \( PE_{gravity} = mgh \), where \( m = 1.20\, \text{kg} \), \( g = 9.81\, \text{m/s}^2 \), and \( h = 0.80\, \text{m} \).Calculate this energy:\[ PE_{gravity} = 1.20 \times 9.81 \times 0.80 = 9.408\, \text{J} \]
04

Set Up and Solve Energy Equation for Spring Compression

The gravitational energy converts into the spring's potential energy. Set \( PE_{spring} = \frac{1}{2} k x^2 \) equal to the gravitational potential energy \( 9.408\, \text{J} \) and solve for \(x\):\[ \frac{1}{2} \times 1600 \times x^2 = 9.408 \]\[ 800x^2 = 9.408 \]\[ x^2 = \frac{9.408}{800} = 0.01176 \]\[ x = \sqrt{0.01176} = 0.1085 \, \text{m} \]The book compresses the spring by approximately \(0.1085\, \text{m}\) or \(10.85\, \text{cm}\).

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Key Concepts

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

Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. When it comes to springs, this energy is in the form of elastic potential energy. Elastic potential energy is stored when a spring is either compressed or stretched. The stored energy can be calculated using the formula:
  • \( PE = \frac{1}{2} k x^2 \)
Here, \(k\) is the spring constant, a measure of the spring's stiffness, and \(x\) is the displacement from the spring's equilibrium (compressed or stretched distance).
If you know the amount of potential energy you want to store, you can rearrange this formula to find out how much the spring should be compressed.
For example, using the given values \(k = 1600 \, \text{N/m}\) and \(PE = 3.20 \, \text{J}\), solve for \(x\) to find the required compression for storing the potential energy.
Spring Compression
Spring compression refers to the amount by which a spring is shortened from its natural length. This is an important concept because it tells us how much a spring must compress to store a certain amount of potential energy.
To find the compression, we use the modified potential energy equation:
  • \( x = \sqrt{\frac{2 \times PE}{k}} \)
In our example, with \(PE = 3.20 \, \text{J}\) and \(k = 1600 \, \text{N/m}\), the spring must compress by \(0.0632 \, \text{m}\) (or \(6.32 \, \text{cm}\)) to store the energy. Spring compression is a key aspect in various real-life applications, such as in shock absorbers in vehicles.
Gravitational Energy
Gravitational energy, or gravitational potential energy, is the energy that an object possesses due to its position in a gravitational field. For instance, an object held at a height above the ground stores gravitational energy. This energy is calculated using:
  • \( PE_{gravity} = mgh \)
Here, \(m\) is the object's mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) near Earth's surface), and \(h\) is the height above a reference point.
In the example of the falling book, the gravitational potential energy before it hits the spring is \(9.408 \, \text{J}\). This energy plays a critical role in understanding how energy is transferred when objects interact with one another.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In the example of the falling book, we see a conversion from gravitational energy to spring potential energy.
When the book is released from a height, its gravitational potential energy transforms into kinetic energy as it falls. Upon impacting the spring, this kinetic energy is converted into the potential energy stored in the spring.
This is a classic illustration of the conservation of energy, where total energy remains the same, merely shifting forms. Understanding energy conversion helps us predict how systems behave and is fundamental in fields like mechanics and engineering.

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Most popular questions from this chapter

A ball is thrown upward with an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

A block with mass \(0.50 \mathrm{~kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of \(0.20 \mathrm{~m}\), as shown in Figure \(7.47 .\) When released, the block moves on a horizontal tabletop for \(1.00 \mathrm{~m}\) before coming to rest. The spring constant \(k\) is \(100 \mathrm{~N} / \mathrm{m} .\) What is the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the block and the tabletop?

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