/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A block of mass \(m=2.0 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 8-39). Find the maximum distance the spring is compressed.

Short Answer

Expert verified
The spring is compressed by approximately 8.94 cm.

Step by step solution

01

Convert Units

First, we need to convert the given height from centimeters to meters for consistency in SI units:Given height, \( h = 40 \text{ cm} = 0.4 \text{ m} \).
02

Calculate Potential Energy at the Top

The gravitational potential energy when the block is released can be calculated using the formula: \[ PE_{top} = mgh \]Where \( m = 2.0 \, \text{kg} \), \( g = 9.8 \frac{\text{m}}{\text{s}^2} \), and \( h = 0.4 \, \text{m} \).So, \[ PE_{top} = 2.0 \, \text{kg} \times 9.8 \, \frac{\text{m}}{\text{s}^2} \times 0.4 \, \text{m} = 7.84 \, \text{J} \].
03

Understand Energy Conservation

By the principle of conservation of energy, the initial gravitational potential energy of the block is converted into spring potential energy at maximum compression. Thus: \[ PE_{top} = PE_{spring} \].
04

Calculate Maximum Spring Compression

The potential energy stored in the spring at maximum compression can be given by: \[ PE_{spring} = \frac{1}{2} k x^2 \]Equating it to the gravitational potential energy: \[ 7.84 = \frac{1}{2} \times 1960 \times x^2 \]Solve for \( x \) to find the maximum compression: \[ x^2 = \frac{7.84 \times 2}{1960} \] \[ x^2 = \frac{15.68}{1960} \] \[ x^2 = 0.008 \] \[ x = \sqrt{0.008} \approx 0.0894 \text{ m} \approx 8.94 \text{ cm} \].

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Key Concepts

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

Conservation of Energy
The concept of conservation of energy is crucial in understanding how energy changes form but is never lost or gained in a closed system. When a block is dropped, its gravitational potential energy at the height will convert into another form of energy as it reaches lower positions. But this conversion follows a balance, according to the principle that energy within a closed system must remain constant.

In the problem, the block's gravitational potential energy (GPE) is converted into spring potential energy as the block compresses the spring. Accessing this concept can help us predict how systems involving energy transformation will behave.

Key points to remember include:
  • Energy can change from one type to another, like from kinetic to potential energy.
  • The total energy of an isolated system remains constant.
  • This principle can predict the outcome once we know the initial conditions and the types of energies involved.
Gravitational Potential Energy
Gravitational potential energy (GPE) is a form of energy related to an object's height and mass. It plays a vital role in energy conversion problems, especially when dealing with dropped objects. Mathematically, GPE is calculated using the formula

\[PE_{top} = mgh\]

where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height.

In our exercise, this concept helps in determining the initial energy when the block is at a given height. With a mass of 2.0 kg and a height of 0.4 m, the gravitational potential energy comes out to be 7.84 Joules. This energy will be transformed as the block moves through its path, compressing the spring.

Understanding GPE helps you predict the energy conversion during motion, especially when working with heights and gravitational effects.
Spring Potential Energy
Spring potential energy comes into play when a block compresses a spring. If you imagine a spring being squished down under pressure, it builds up energy which can "bounce back" when the pressure is released. This is formalized in Hooke's Law, which says that the energy stored in a compressed or stretched spring is given by the formula

\[PE_{spring} = \frac{1}{2}kx^2\]

where \(k\) is the spring constant and \(x\) is the displacement of the spring from its rest position.

In our exercise, finding the spring potential energy when the spring is compressed helps us understand how much potential energy the block transforms into at the spring's furthest compression point. Solving the equation
\[7.84 = \frac{1}{2} \times 1960 \times x^2\]
leads us to determine the maximum compression of the spring, which is approximately 0.0894 m or 8.94 cm.

Knowing how spring potential energy works helps solve complex problems in contexts like mechanics and design where springs are pivotal.

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Most popular questions from this chapter

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m} .\) The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0 \mathrm{~cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

The temperature of a plastic cube is monitored while the cube is pushed \(3.0 \mathrm{~m}\) across a floor at constant speed by a horizontal force of \(15 \mathrm{~N}\). The thermal energy of the cube increases by \(20 \mathrm{~J} .\) What is the increase in the thermal energy of the floor along which the cube slides?

A \(1500 \mathrm{~kg}\) car begins sliding down a \(5.0^{\circ}\) inclined road with a speed of \(30 \mathrm{~km} / \mathrm{h}\). The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled \(50 \mathrm{~m}\) along the road, its speed is \(40 \mathrm{~km} / \mathrm{h}\). (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

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