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

Short Answer

Expert verified
(a) 0.0632 m; (b) 0.108 m.

Step by step solution

01

Understanding the Formula for Potential Energy in a Spring

The potential energy stored in a compressed or stretched spring can be calculated using the formula: \[ U = \frac{1}{2} k x^2 \]where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression distance from its equilibrium position.
02

Calculate Compression for Stored Energy

To find the distance \( x \) for which the spring stores 3.20 J of energy, set \( U = 3.20 \) J and \( k = 1600 \) N/m in the potential energy formula: \[ 3.20 = \frac{1}{2} \times 1600 \times x^2 \]Solve for \( x \): \[ x^2 = \frac{3.20 \times 2}{1600} = \frac{6.40}{1600} \]\[ x^2 = 0.004 \]\[ x = \sqrt{0.004} \approx 0.0632 \text{ m} \]
03

Calculate Initial Mechanical Energy for Dropped Book

The potential energy when the book is at height 0.80 m is given by:\[ PE = mgh = 1.20 \times 9.8 \times 0.80 \]\[ PE = 9.408 \text{ J} \]This is the potential energy due to gravity before the book hits the spring.
04

Considering Energy Conservation

When the book compresses the spring to its maximum point, all gravitational potential energy is converted into spring potential energy: \[ \frac{1}{2} k x^2 = mgh \]From step 3, \( mgh = 9.408 \text{ J} \). Already known \( k = 1600 \text{ N/m} \). Solve for \( x \): \[ 9.408 = \frac{1}{2} \times 1600 \times x^2 \]\[ 9.408 = 800x^2 \]\[ x^2 = \frac{9.408}{800} \]\[ x = \sqrt{0.01176} \approx 0.108 \text{ m} \]
05

Conclusion for Both Part A and Part B

For part (a), the spring must be compressed about 0.0632 m to store 3.20 J of energy. For part (b), the maximum compression distance of the spring due to the drop of the book is approximately 0.108 m.

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Key Concepts

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

Force Constant
In the study of springs, the term 'force constant', or spring constant, denoted by \( k \), is pivotal. This constant represents the stiffness of a spring. A higher \( k \) value implies a stiffer spring that requires more force to compress or stretch it by a certain length. The unit of \( k \) is Newton per meter (N/m). An understanding of the force constant is crucial when calculating how much potential energy a spring can store when compressed or stretched.

Consider the formula for potential energy in a spring:
  • \( U = \frac{1}{2} k x^2 \)
Here, \( U \) stands for potential energy, while \( x \) is the displacement of the spring from its equilibrium position. The equation shows that energy stored in the spring depends on both the force constant and the square of the displacement. In practical applications, this means knowing \( k \) allows you to predict how much energy will be stored for a given compression or stretch, which is crucial in designing systems that use spring mechanics efficiently.
Energy Conservation
Energy conservation is a fundamental principle in physics which states that energy cannot be created or destroyed but only transformed from one form to another. In this exercise, energy conservation becomes evident when a book is dropped onto a spring. Initially, the book possesses gravitational potential energy due to its height above the ground.

As the book falls, this gravitational potential energy is converted into spring potential energy at the point of maximum compression. The equation that reflects this conversion is:
  • \( mgh = \frac{1}{2} k x^2 \)
Here, \( mgh \) represents the initial gravitational potential energy, while \( \frac{1}{2} k x^2 \) symbolizes the energy stored in the spring. By setting these two equations equal, you can solve for \( x \), finding the spring's compression distance that matches the book's potential energy loss. This principle is applicable in various practical situations, ensuring efficient design, from amusement park rides to engineering shock absorption systems.
Mechanical Energy
Mechanical energy comprises both potential and kinetic energy in a system. In the context of the spring problem, we focus primarily on potential energy. Initially, the book's position contributes gravitational potential energy, calculated as \( mgh \). Upon impact with the spring, this energy is transformed into another form of potential energy stored within the spring.

The conservation of mechanical energy involves maintaining the total energy of the system constant as long as no external forces, like friction, interfere. This is why understanding mechanical energy components is so vital.
  • Potential Energy in Gravity: \( PE = mgh \)
  • Potential Energy in a Spring: \( PE = \frac{1}{2} k x^2 \)
These formulas highlight how mechanical energy changes form but remains constant in quantity, providing a predictable and calculable way to assess energy transfers in systems. This principle helps engineers and scientists design systems where energy transitions smoothly, yielding functionality such as energy-efficient springs or controlled motion mechanics.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A 20.0 kg rock slides on a rough horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average thermal power is produced as the rock stops?

\(\bullet\) (a) How many joules of energy does a 100 watt lightbulb use every hour? (b) How fast would a 70 kg person have to run to have that amount of kinetic energy? Is it possible for a per- son to run that fast? (c) How high a tree would a 70 kg person have to climb to increase his gravitational potential energy rela- tive to the ground by that amount? Are there any trees that tall?

\(\bullet\) \(\bullet\) A 250 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35 \(\mathrm{N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is com- pressed 12.0 \(\mathrm{cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

\(\bullet\) \(\bullet\) An elevator has mass \(600 \mathrm{kg},\) not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 \(\mathrm{m}\) (five floors) in \(16.0 \mathrm{s},\) and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 \(\mathrm{kg}\) .

\(\bullet\) The speed of hailstones. Although the altitude may vary considerably, hailstones sometimes originate around 500 \(\mathrm{m}\) (about 1500 \(\mathrm{ft} )\) above the ground. (a) Neglecting air drag, how fast will these hailstones be moving when they reach the ground, assuming that they started from rest? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in mph. (b) From your own experience, are hailstones actually falling that fast when they reach the ground? Why not? What has happened to most of the initial potential energy?

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