/*! 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 131 131 Fasten one end of a vertical... [FREE SOLUTION] | 91Ó°ÊÓ

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131 Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the spring's potential energy.

Short Answer

Expert verified
The loss of gravitational potential energy equals twice the spring's energy gain.

Step by step solution

01

Understand the Problem

The problem involves attaching a cabbage to a vertical spring. We need to analyze the energy changes as the cabbage is slowly lowered until the forces are balanced. Our goal is to show that the loss in the gravitational potential energy is twice the gain in the spring's potential energy.
02

Define Gravitational Potential Energy Loss

The gravitational potential energy (GPE) lost by the cabbage as it moves a distance \( d \) downward is given by the formula \( \Delta U_g = m g d \), where \( m \) is the mass of the cabbage, \( g \) is the acceleration due to gravity, and \( d \) is the displacement.
03

Define Spring Potential Energy Gain

The potential energy stored in the spring, when it is stretched by a distance \( x \), is \( \Delta U_s = \frac{1}{2} k x^2 \), where \( k \) is the spring constant. At equilibrium, the spring force equals the gravitational force, so \( k x = m g \).
04

Relate Displacement and Stretch

Since the cabbage is slowly lowered to the equilibrium position, the displacement \( d \) of the cabbage is the same as the stretch \( x \) of the spring, i.e., \( d = x \).
05

Calculate Total Gravitational Potential Energy Loss

With \( d = x \), the total gravitational potential energy lost is \( \Delta U_g = m g x \).
06

Substitute Stretch into the Spring Energy Gain

Substitute \( x = \frac{m g}{k} \) into the spring potential energy formula, giving \( \Delta U_s = \frac{1}{2} k \left( \frac{m g}{k} \right)^2 = \frac{m^2 g^2}{2k} \).
07

Compare Energies

From equilibrium condition, \( m g x = k x^2 \) so \( x = \frac{m g}{k} \). Thus, \( m g d = 2 \cdot \frac{1}{2} k x^2 \), showing that \( \Delta U_g = 2 \Delta U_s \), which confirms that the loss of gravitational potential energy is twice the gain in spring potential energy.

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Key Concepts

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

Gravitational Potential Energy
When we hear the term "gravitational potential energy," it refers to the energy stored in an object due to its position in a gravitational field. In simpler terms, it's the energy an object has because of where it is placed—typically, how high it is above the ground. The formula to calculate this energy is \( \Delta U_g = m g d \) where:
  • \( m \) represents the mass of the object (in this scenario, the cabbage),
  • \( g \) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth), and
  • \( d \) denotes the vertical distance the object moves.
When the cabbage is lowered, it loses gravitational potential energy, which we can think of as the energy it could have used to fall. The amount of energy lost depends directly on the distance it falls and its mass.
Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is either compressed or stretched. Think of a spring as a sort of energy bank account that stores mechanical energy based on its change from its original shape. This potential energy is given by the formula:\[ \Delta U_s = \frac{1}{2} k x^2 \]where:
  • \( k \) is the spring constant that tells us how stiff the spring is, and
  • \( x \) is the distance the spring is stretched or compressed from its natural length.
In the problem, as the cabbage stretches the spring, the spring accumulates potential energy. This energy is dependent on the square of the displacement, emphasizing that even small stretches can lead to significant energy storage. At the equilibrium position, the energy gained by the spring is a direct consequence of this stretching.
Equilibrium in Mechanics
Equilibrium refers to a state where all forces acting on an object are perfectly balanced, meaning there is no net force causing the object to move. In mechanics, equilibrium can be either static or dynamic, although in this problem, we deal with static equilibrium.When the cabbage is gently lowered, it compresses the spring until the upward force from the spring equals the downward gravitational force. This balance of forces is represented by \[ k x = m g \] indicating that the force exerted by the spring (\( k x \)) matches the weight of the cabbage (\( m g \)).In such situations, understanding equilibrium helps us relate the changes in gravitational and spring potential energies. Since energy is conserved, the energy lost by the cabbage gets transferred to the spring, albeit with some interesting proportional relationships, as shown—the loss in gravitational potential energy is twice the gain in spring potential energy. This clever conservation and distribution of energy are core principles in physics, illustrating how systems naturally balance forces and energies.

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Most popular questions from this chapter

A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. \(8-37\) ) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

Next, the particle is released from rest at \(x=0 .\) What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem \(81,\) a \(2.00 \mathrm{~kg}\) particle is released at \(x=5.00 \mathrm{~m}\) with an initial velocity of \(3.45 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis. (a) If the particle can reach \(x=0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive \(x\) direction when it is released at \(x=5.00 \mathrm{~m}\) at speed \(3.45 \mathrm{~m} / \mathrm{s} .\) (b) If the particle can reach \(x=13.0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point?

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is 0.10 . (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s},\) what is her speed at the bottom?

Two blocks, of masses \(M=2.0 \mathrm{~kg}\) and \(2 M,\) are connected to a spring of spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) that has one end fixed, as shown in Fig. \(8-69 .\) The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen \(0.090 \mathrm{~m} ?\) (b) What is the kinetic energy of the hanging block when it has fallen that \(0.090 \mathrm{~m} ?\) (c) What maximum distance does the hanging block fall before momentarily stopping?

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