/*! 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 18 Stone Rests on Spring shows an \... [FREE SOLUTION] | 91Ó°ÊÓ

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Stone Rests on Spring shows an \(8.00 \mathrm{~kg}\) stone at rest on a spring. The spring is compressed \(10.0 \mathrm{~cm}\) by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional \(30.0\) \(\mathrm{cm}\) and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?

Short Answer

Expert verified
The spring constant is 784 N/m. The elastic potential energy is 62.72 J. The change in gravitational potential energy is 62.72 J. The maximum height from the release point is 1.10 m.

Step by step solution

01

Determine the Spring Constant (k)

Given:- Mass of the stone, m = 8.00 kg- Compression of the spring, x = 10.0 cm = 0.10 mThe force due to the weight of the stone (F) is given by \( F = mg \)where g is the acceleration due to gravity (\( 9.8 \, m/s^2 \)).Calculate the force:\( F = 8.00 \, kg \times 9.8 \, m/s^2 = 78.4 \, N \)The spring constant k can be found using Hooke's Law:\( F = kx \)Solving for k:\( k = \frac{F}{x} = \frac{78.4 \, N}{0.10 \, m} = 784 \, N/m \)
02

Calculate the Elastic Potential Energy (U)

Given:- Additional compression, x = 30.0 cm = 0.30 m- Total compression = initial compression + additional compression = 0.10 m + 0.30 m = 0.40 m- Spring constant k from Step 1 = 784 N/mThe elastic potential energy stored in the spring is given by:\( U = \frac{1}{2}kx^2 \)Calculate U:\( U = \frac{1}{2} \times 784 \, N/m \times (0.40 \, m)^2 = \frac{1}{2} \times 784 \times 0.16 = 62.72 \, J \)
03

Determine Change in Gravitational Potential Energy (\Delta U_g\)

Given:- Mass of the stone, m = 8.00 kg- Additional compression, x = 0.30 m- g = 9.8 m/s^2Change in gravitational potential energy when the stone moves from the release point to its maximum height (where all elastic potential energy converts to gravitational potential energy) is given by:\( \Delta U_g = mgh \)Since elastic potential energy converts to gravitational potential energy:\( \Delta U_g = U \)From Step 2, U = 62.72 JSo, calculate h using:\( 62.72 J = 8.00 kg \times 9.8 m/s^2 \times h \)Solving for h:\( h = \frac{62.72}{8.00 \times 9.8} = 0.80 \, m \)
04

Find Maximum Height (h) From Release Point

The maximum height measured from the release point (additional 0.30 m compression) is given by summing the distance the stone moves upwards and the initial 0.30 m compression.Since the stone travels upwards by h = 0.80 m from Step 3:Total height from the release point is:\( h_{total} = 0.80 m + 0.30 m = 1.10 m \)

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Key Concepts

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

Hooke's Law
Hooke's Law describes the behavior of springs. It states that the force needed to compress or extend a spring by some distance is proportional to that distance. Mathematically, it is represented as:
\( F = kx \)
Here, \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement. In this exercise, the weight of the stone compressed the spring, and we used Hooke's Law to find the spring constant. By rearranging the formula to \( k = \frac{F}{x} \), and substituting the known values, we calculated the spring constant as 784 N/m. Understanding Hooke's Law is crucial for solving problems involving springs. It shows that the force required to compress a spring increases linearly with the distance compressed.
Elastic Potential Energy
Elastic potential energy is the energy stored in an elastic object, like a spring, when it is compressed or stretched. The formula for elastic potential energy is:
\( U = \frac{1}{2}kx^2 \)
Here, \( U \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the compression or extension distance. In our exercise, the spring was further compressed by 0.30 m, making the total compression 0.40 m. We used the spring constant calculated earlier and plugged in the values to find the elastic potential energy. By substituting \( k = 784 \) N/m and \( x = 0.40 \) m in the formula, we got the elastic potential energy as 62.72 J. This concept helps us understand how energy is stored and transferred within a spring system.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is given by the formula:
\( U_g = mgh \)
In this formula, \( U_g \) represents gravitational potential energy, \( m \) is mass, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height. In our problem, we calculated the change in gravitational potential energy by determining the height the stone would rise after the spring is released. We equated the elastic potential energy (62.72 J) to the change in gravitational potential energy and solved for \( h \). This gave us a height of 0.80 m. This conversion between energy forms shows the principle of conservation of energy.
Spring Compression
Spring compression is the act of pressing a spring together, reducing its length. The amount of compression affects the force stored and the potential energy within the spring. When the stone was placed on the spring, it compressed initially by 0.10 m. This displacement was crucial to calculating the spring constant using Hooke’s Law. Further, the stone was compressed an additional 0.30 m, making the total compression 0.40 m. This total compression was used to determine the elastic potential energy before the stone was released. Understanding the concepts of compression and displacement is essential for solving problems involving springs.
Physics Problem-Solving
Physics problem-solving often involves multiple steps and the application of different principles. Here's a structured approach we used for this exercise:
  • Identify given values and what needs to be found.
  • Use fundamental laws and formulas, like Hooke's Law for spring constant.
  • Calculate the required quantities step-by-step, such as spring constants and potential energies.
  • Apply energy conservation principles to find relationships between different forms of energy.
  • Recheck units and calculations to ensure accuracy.
This structured method helps break down complex problems into manageable steps, leading to accurate solutions. Practice and familiarity with key concepts enhance problem-solving skills in physics.

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Most popular questions from this chapter

Worker Pushes Block A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

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. Frisbee A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m}\), its speed is \(10.5 \mathrm{~m} / \mathrm{s}\). What was the reduction in the mechanical energy of the Frisbee- Earth system because of air drag?

Jumping into a Haystack Tom Sawyer wanders out to the barn one fine summer's day. He notices that a haystack has recently been built just outside the barn. The barn has a second-story door into which the hay will be hauled into the barn by a crane. Tom decides it would be a neat idea to jump out of the second-story door onto the haystack. However, he knows from sad experience that if he jumps out of the second-story door onto the ground, that he is likely to break his leg. Knowing lots of physics, Tom decides to estimate whether the haystack will be able to break his fall. He estimates the height of the haystack to be 3 meters. He presses down on top of the stack and discovers that to compress the stack by \(25 \mathrm{~cm}\), he has to exert a force of about \(50 \mathrm{~N}\). The barn door is 6 meters above the ground. Solve the problem by breaking it into pieces as follows: 1\. Model the haystack by a spring. What is its spring constant? 2\. Is the haystack tall enough to bring his speed to zero? (Estimate using conservation of energy.) 3\. If he does come to a stop before he hits the ground, what will the average force exerted on him be?

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

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