/*! 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 6 You use your hand to stretch a s... [FREE SOLUTION] | 91Ó°ÊÓ

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You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

Short Answer

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a) The spring's ΔU is positive b) The spring's ΔU is negative c) The hand's ΔU is positive d) The hand's ΔU is negative e) None of the above statements is true Answer: b) The spring's ΔU is negative and d) The hand's ΔU is negative

Step by step solution

01

Determine potential energy changes in the spring

In the given situation, the spring is first stretched to a displacement \(x\) and then brought back to its equilibrium position. Let's calculate the potential energy change during this process: Initial potential energy, \(U_i = \frac{1}{2} kx^2\) Final potential energy, \(U_f = 0\), because the spring is back to its equilibrium position. Now, calculate the change in potential energy, \(\Delta U = U_f - U_i\). \(\Delta U_{spring} = 0 - \frac{1}{2} kx^2 = - \frac{1}{2} kx^2\) For the spring, \(\Delta U_{spring} < 0\), which means the potential energy of the spring has decreased during the process.
02

Determine potential energy changes in the hand

The work done by the hand to stretch the spring and bring it back to its equilibrium position is equal to the change in potential energy in the spring. In this case, the work done by the hand is given by: \(W_{hand} = -\Delta U_{spring}\) \(W_{hand} = -(- \frac{1}{2} kx^2)\) \(W_{hand} = \frac{1}{2} kx^2\) Since work done by the hand is positive, it means that the potential energy of the hand has decreased during the process. Therefore, \(\Delta U_{hand} < 0\).
03

Analyze the options

Based on the analysis in the previous steps, let's go through each option: a) The spring's \(\Delta U\) is positive - False, because \(\Delta U_{spring} < 0\). b) The spring's \(\Delta U\) is negative - True, as \(\Delta U_{spring} = -\frac{1}{2} kx^2\). c) The hand's \(\Delta U\) is positive - False, because \(\Delta U_{hand} < 0\). d) The hand's \(\Delta U\) is negative - True, as the hand's potential energy has decreased during the process. e) None of the above statements is true - False, because options b) and d) are true. Therefore, the correct answer is that the spring's \(\Delta U\) is negative (option b) and the hand's \(\Delta U\) is negative (option d).

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Key Concepts

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

Spring Potential Energy
Understanding spring potential energy is key to grasping various physics concepts. Imagine a spring at rest. When you apply a force to stretch or compress it, you're storing energy in the form of spring potential energy. Mathematically, it's given by the equation \( U = \frac{1}{2} kx^2 \), where \( k \) represents the spring constant, indicating the stiffness of the spring, and \( x \) is the displacement from its equilibrium position.

In the context of the exercise, when you stretch the spring to displacement \( x \) and then allow it to return to the equilibrium, the stored potential energy is transformed back into other forms, usually kinetic energy or work done by the spring's force. The concept that the energy stored in the spring depends on the square of the displacement is essential, meaning even small stretches or compressions can store significant amounts of energy within stiff springs. This fundamental principle is widely used in mechanics, from watches to car suspensions.
Work-Energy Principle
The work-energy principle is a cornerstone in physics, providing a clear link between the forces acting on an object and its energy. In simplest terms, it states that work done by forces on an object results in a change in that object's kinetic energy. Mathematically, \( W = \Delta KE \), where \( W \) is the work and \( KE \) is the kinetic energy. When looking at springs, this principle can be extended; the work done in stretching or compressing a spring is manifested as potential energy within the spring.

For the exercise, when you use your hand to stretch and then release the spring, your hand is doing work against the spring force. This work is initially stored as potential energy in the spring and is then released as the spring returns to equilibrium. The work-energy principle reminds us that energy is neither created nor destroyed but simply transformed from one type to another, such as from potential to kinetic energy or vice-versa.
Mechanical Energy Conservation
The conservation of mechanical energy is a fundamental concept that states when only conservative forces (like gravity and spring forces) act on a system, the total mechanical energy remains constant. This total includes both kinetic energy (\( KE \) - energy of motion) and potential energy (\( PE \) - stored energy).

In the exercise we've been considering, even as the spring goes from stretched back to its equilibrium position, the total mechanical energy of the system remains the same (if we ignore non-conservative forces like friction). Energy transformations occur within the system: Potential energy (spring potential energy when stretched) converts into kinetic energy (motion) and potentially into work (done by the hand), but the total amount of mechanical energy stays constant throughout the process.

The mechanical energy conservation principle helps explain why the spring returns to its original position after being disturbed – it's an interplay of energy conversion within a conservative force field, with no net loss in the system's mechanical energy.

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Most popular questions from this chapter

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

A package is dropped on a horizontal conveyor belt. The mass of the package is \(m,\) the speed of the conveyor belt is \(v\), and the coefficient of kinetic friction between the package and the belt is \(\mu_{\mathrm{k}}\) a) How long does it take for the package to stop sliding on the belt? b) What is the package's displacement during this time? c) What is the energy dissipated by friction? d) What is the total work supplied by the system?

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

Calculate the force \(F(y)\) associated with each of the following potential energies: a) \(U=a y^{3}-b y^{2}\) b) \(U=U_{0} \sin (c y)\)

Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 10.0 -kg mass is released and falls through a vertical distance of \(1.00 \mathrm{~m}\) before hitting the ground. Use conservation of mechanical energy to determine: a) how fast the 5.00 -kg mass is moving just before the 10.0 -kg mass hits the ground; and b) the maximum height attained by the 5.00 -kg mass.

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