/*! 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 \(h=50 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 8-27). Find the maximum distance the spring is compressed.

Short Answer

Expert verified
The maximum distance the spring is compressed is 0.1 meters (or 10 cm).

Step by step solution

01

Analyze the Potential Energy of the Block

The block has gravitational potential energy when it is at height \(h\). We calculate this energy using the formula: \( PE = mgh \), where \( m = 2.0 \mathrm{~kg} \), \( g = 9.8 \mathrm{~m/s^2} \), and \( h = 0.50 \mathrm{~m} \). Thus, the potential energy is \( PE = 2.0 \times 9.8 \times 0.50 = 9.8 \mathrm{~J} \).
02

Set up the Conservation of Energy Equation

The gravitational potential energy of the block is transformed into the elastic potential energy of the spring. When the spring is at its maximum compression, its potential energy is given by \( \text{Elastic Potential Energy} = \frac{1}{2} k x^2 \), where \( x \) is the compression distance and \( k = 1960 \mathrm{~N/m} \). Thus, \( 9.8 = \frac{1}{2} \times 1960 \times x^2 \).
03

Solve for the Compression Distance

Rearrange the equation to solve for \( x^2 \): \( 9.8 = 980 x^2 \).\( x^2 = \frac{9.8}{980} \). Calculating this gives \( x^2 = 0.01 \). Then, take the square root of both sides to find \( x \): \( x = \sqrt{0.01} = 0.1 \mathrm{~m} \).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It depends on three key factors: the mass of the object, the gravitational force acting on it, and the height from which it is dropped. In mathematical terms, it is expressed by the formula:
  • \( PE = mgh \)
where:
  • \( m \) is the mass of the object in kilograms,
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth),
  • \( h \) is the height in meters from which the object is dropped.

In the given exercise, understanding gravitational potential energy is critical as it is the initial form of energy the block possesses. When the block drops, this energy transforms completely into another form of energy when it impacts the spring. This is exactly what makes potential energy a dynamic component in the context of energy conservation, serving as a measurable bridge in the study of motion and force.
Elastic Potential Energy
Elastic potential energy is the potential energy stored in elastic materials, such as springs, when they are stretched or compressed. The formula for elastic potential energy in a spring is:
  • \( ext{Elastic Potential Energy} = \frac{1}{2} k x^2 \)
where:
  • \( k \) is the spring constant in newtons per meter (N/m), indicating the spring's stiffness,
  • \( x \) is the displacement or compression of the spring from its equilibrium position in meters.

When the block lands on the spring, its gravitational potential energy converts into elastic potential energy, causing the spring to compress. The amount of energy stored is proportional to the square of the compression distance, making the relationship quadratic. This means a small increase in compression leads to a larger increase in elastic potential energy, demonstrating how energy conservation principles apply in practical scenarios, such as cushioning systems, car suspensions, and more.
Spring Compression
Spring compression refers to the degree to which a spring is compacted by an external force. It’s a measure of the energy transformation from gravitational to elastic potential energy. In this exercise, the fundamental goal is to determine the maximum compression of the spring when the block lands on it.
  • Using energy conservation, the gravitational potential energy of the block is fully transformed into the elastic potential energy of the spring.
  • The conversion allows us to solve for compression with the formula relating the energies: \( 9.8 = \frac{1}{2} \times 1960 \times x^2 \).
  • By solving the equation, we can derive the compression distance: \( x = \sqrt{0.01} = 0.1 \, \text{meters} \).

This compression calculation is not just a mathematical exercise, but also provides insights into how springs are designed to absorb impact, efficiently converting motion and potential energy into a form that slows down and eventually stops moving objects safely.

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Most popular questions from this chapter

A loaded truck of mass \(3000 \mathrm{~kg}\) moves on a level road at a constant speed of \(6.000 \mathrm{~m} / \mathrm{s}\). The frictional force on the truck from the road is \(1000 \mathrm{~N}\). Assume that air drag is negligible. (a) How much work is done by the truck engine in \(10.00 \mathrm{~min}\) ? (b) After \(10.00 \mathrm{~min}\), the truck enters a hilly region whose inclination is \(30^{\circ}\) and continues to move with the same speed for another \(10.00 \mathrm{~min}\). What is the total work done by the engine during that period against the gravitational force and the frictional force? (c) What is the total work done by the engine in the full \(20 \mathrm{~min}\) ?

A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(27 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 22 \(\mathrm{m} / \mathrm{s}\), landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. Your friend's outstretchedhands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-19). (a) How much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released, and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U\), (g) \(U\) at the release point, and (h) \(U\) at her hands.

A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(11.5 \mathrm{~m}\) high. The snowball's initial velocity is \(16.0 \mathrm{~m} / \mathrm{s}\), directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

Tarzan, who weighs \(688 \mathrm{~N}\), swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long (Fig. 8-28). From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\) (a) Does the vine break? (b) If it does not, what is the greatest force on it during the swing? If it does, at what angle with the vertical does it break?

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