/*! 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 20 A 1.20 -kg piece of cheese is pl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 1.20 -kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{N} / \mathrm{m}\) that is compressed 15.0 \(\mathrm{cm}\) . When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

Short Answer

Expert verified
The cheese rises 1.72 meters from the initial position.

Step by step solution

01

Understand the Concept

The problem involves energy conservation where the potential energy stored in the spring gets converted into the gravitational potential energy of the cheese. We will use the concepts of spring potential energy and gravitational potential energy.
02

Determine Initial Spring Potential Energy

The potential energy in the spring when it is compressed is given by the formula \( PE_{spring} = \frac{1}{2} k x^2 \), where \( k = 1800\, \text{N/m} \) and \( x = 0.15\, \text{m} \). So, \( PE_{spring} = \frac{1}{2} \times 1800 \times (0.15)^2 \).
03

Set Up Energy Conservation Equation

At the maximum height reached by the cheese, all the spring potential energy will convert into gravitational potential energy. \( PE_{gravitational} = mgh \), where \( m = 1.20\, \text{kg} \), \( g = 9.8\, \text{m/s}^2 \), and \( h \) is the height raised.
04

Calculate the Spring's Initial Potential Energy

Calculate \( PE_{spring} = \frac{1}{2} \times 1800 \times (0.15)^2 = 20.25 \, \text{J} \).
05

Solve for Maximum Height

Using the energy conservation equation \( mgh = PE_{spring} \). Substitute \( h = \frac{PE_{spring}}{mg} \), so \( h = \frac{20.25}{1.20 \times 9.8} = 1.72 \, \text{m} \).
06

Result Interpretation

The cheese rises to a maximum height of 1.72 meters above the initial compressed position of the spring.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched from its natural length. This energy is a form of mechanical energy that follows Hooke's Law, which states that the force required to compress or extend a spring is directly proportional to the distance it is stretched. To calculate the spring potential energy (\( PE_{spring} \)), use the formula: \[ PE_{spring} = \frac{1}{2} k x^2 \]where:
  • \( k \) is the force constant (or stiffness) of the spring measured in newtons per meter (N/m),
  • \( x \) is the displacement from the spring's original position in meters (m).
In the given exercise, a compressed spring with a force constant of 1800 N/m and a displacement of 0.15 m stores energy. Plugging these values into the formula gives us the stored spring potential energy, which eventually helps the cheese rise into the air when the spring is released.
Understanding this principle helps us see how potential energy in a spring can convert into other forms of energy like gravitational potential energy.
Gravitational Potential Energy
Gravitational potential energy (\( PE_{gravitational} \)) is the energy an object possesses due to its position in a gravitational field. This energy increases when an object is lifted to a greater height above the ground. In the realm of physics, it is important to realize that this form of energy plays a crucial role in the conservation of mechanical energy. The formula to determine gravitational potential energy is:\[ PE_{gravitational} = mgh \] where:
  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity, usually approximated as 9.8 m/s² on the surface of the Earth,
  • \( h \) is the height above the reference point (in meters).
In our exercise, the cheese rises to a height determined by converting the spring's stored energy entirely into gravitational potential energy, illustrating the conservation of energy in action.
This concept is fundamental to understanding how potential energy can be converted into kinetic energy and vice versa, playing a significant role in various physical phenomena.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in any given system. A vital concept in physics, mechanical energy demonstrates the interplay and conversion between potential energy forms, such as spring and gravitational, as well as kinetic energy. The conservation of mechanical energy principle states that in a closed system with no external forces, the total mechanical energy remains constant.For systems involving springs and gravitational elements, this is often expressed as:\[ \text{Total Mechanical Energy} = KE + PE_{spring} + PE_{gravitational} \]Understanding mechanical energy is key to analyzing many physical systems, such as the motion of the cheese in our exercise. Here, as the spring releases, the potential energy stored in the spring converts into kinetic energy as the cheese moves upward. At the peak of its trajectory, the kinetic energy transitions entirely into gravitational potential energy.
Knowing how mechanical energy works allows students to appreciate how energy transforms and conserves in different situations, illuminating the beauty and power of physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He sturts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance \(h \leq d\) above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special cases of \(h=d\) and \(h=0 ?\) (b) Find a numerical value for the average friction force a 75 -kg fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m}\) (c) In terms of \(g, h,\) and \(d,\) what is the speed of the fireman when he is a distance \(y\) above the bottom of the pole?

CALE An object moving in the \(x y\) -plane is acted on by a conservative force described by the potential-energy function \(U(x, y)=\alpha\left(1 / x^{2}+1 / y^{2}\right),\) where \(\alpha\) is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{i}\) and \(\hat{J} .\)

On a horizontal surface, a crate with mass 50.0 \(\mathrm{kg}\) is placed against a spring that stores 360 \(\mathrm{J}\) of energy. The spring is released, and the crate slides 5.60 \(\mathrm{m}\) before coming to rest. What is the speed of the crate when it is 2.00 \(\mathrm{m}\) from its initial position?

CALC A cutting tool under microprocessor control has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{J}}\) , a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}\) .\(y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if this displacement is along the straight line \(y=x\) that connects these two points. (b) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{\boldsymbol{F}}\) along these two paths. Is \(\vec{\boldsymbol{F}}\) conservative or nonconservative? Explain.

CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.