Skip to Content
OpenStax Logo
University Physics Volume 1

8.1 Potential Energy of a System

University Physics Volume 18.1 Potential Energy of a System
Buy book
  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Relate the difference of potential energy to work done on a particle for a system without friction or air drag
  • Explain the meaning of the zero of the potential energy function for a system
  • Calculate and apply the gravitational potential energy for an object near Earth’s surface and the elastic potential energy of a mass-spring system

In Work, we saw that the work done on an object by the constant gravitational force, near the surface of Earth, over any displacement is a function only of the difference in the positions of the end-points of the displacement. This property allows us to define a different kind of energy for the system than its kinetic energy, which is called potential energy. We consider various properties and types of potential energy in the following subsections.

Potential Energy Basics

In Motion in Two and Three Dimensions, we analyzed the motion of a projectile, like kicking a football in Figure 8.2. For this example, let’s ignore friction and air resistance. As the football rises, the work done by the gravitational force on the football is negative, because the ball’s displacement is positive vertically and the force due to gravity is negative vertically. We also noted that the ball slowed down until it reached its highest point in the motion, thereby decreasing the ball’s kinetic energy. This loss in kinetic energy translates to a gain in gravitational potential energy of the football-Earth system.

As the football falls toward Earth, the work done on the football is now positive, because the displacement and the gravitational force both point vertically downward. The ball also speeds up, which indicates an increase in kinetic energy. Therefore, energy is converted from gravitational potential energy back into kinetic energy.

An illustration of a football’s trajectory and energy. The kicker kicks the ball, doing work on it and giving it maximum kinetic energy. The potential energy is minimum. This is point one. On the way up, at point two, the kinetic energy of the ball decreases and its potential energy decreases. At the highest point, point three, the kinetic energy of the ball  is at its minimum and its potential energy is maximum. As the ball descends, point four, the kinetic energy increases and the potential energy decreases. The receiver catches the ball at the same height above the ground as it was kicked, at point five. The kinetic energy equals maximum, potential energy is minimum.
Figure 8.2 As a football starts its descent toward the wide receiver, gravitational potential energy is converted back into kinetic energy.

Based on this scenario, we can define the difference of potential energy from point A to point B as the negative of the work done:

ΔUAB=UBUA=WAB.ΔUAB=UBUA=WAB.
(8.1)

This formula explicitly states a potential energy difference, not just an absolute potential energy. Therefore, we need to define potential energy at a given position in such a way as to state standard values of potential energy on their own, rather than potential energy differences. We do this by rewriting the potential energy function in terms of an arbitrary constant,

ΔU=U(r)U(r0).ΔU=U(r)U(r0).
(8.2)

The choice of the potential energy at a starting location of r0r0 is made out of convenience in the given problem. Most importantly, whatever choice is made should be stated and kept consistent throughout the given problem. There are some well-accepted choices of initial potential energy. For example, the lowest height in a problem is usually defined as zero potential energy, or if an object is in space, the farthest point away from the system is often defined as zero potential energy. Then, the potential energy, with respect to zero at r0,r0, is just U(r).U(r).

As long as there is no friction or air resistance, the change in kinetic energy of the football equals the change in gravitational potential energy of the football. This can be generalized to any potential energy:

ΔKAB=ΔUAB.ΔKAB=ΔUAB.
(8.3)

Let’s look at a specific example, choosing zero potential energy for gravitational potential energy at convenient points.

Example 8.1

Basic Properties of Potential Energy A particle moves along the x-axis under the action of a force given by F=ax2F=ax2, where a=3N/m2a=3N/m2. (a) What is the difference in its potential energy as it moves from xA=1mxA=1m to xB=2mxB=2m? (b) What is the particle’s potential energy at x=1mx=1m with respect to a given 0.5 J of potential energy at x=0x=0?

Strategy(a) The difference in potential energy is the negative of the work done, as defined by Equation 8.1. The work is defined in the previous chapter as the dot product of the force with the distance. Since the particle is moving forward in the x-direction, the dot product simplifies to a multiplication (i^·i^=1i^·i^=1). To find the total work done, we need to integrate the function between the given limits. After integration, we can state the work or the change in potential energy. (b) The potential energy function, with respect to zero at x=0x=0, is the indefinite integral encountered in part (a), with the constant of integration determined from Equation 8.3. Then, we substitute the x-value into the function of potential energy to calculate the potential energy at x=1m.x=1m.

Solution

  1. The work done by the given force as the particle moves from coordinate x to x+dxx+dx in one dimension is
    dW=F·dr=Fdx=ax2dx.dW=F·dr=Fdx=ax2dx.

    Substituting this expression into Equation 8.1, we obtain
    ΔU=W=x1x2ax2dx=13(3N/m2)x3|1m2m=7J.ΔU=W=x1x2ax2dx=13(3N/m2)x3|1m2m=7J.
  2. The indefinite integral for the potential energy function in part (a) is
    U(x)=13ax3+const.,U(x)=13ax3+const.,

    and we want the constant to be determined by
    U(0)=0.5J.U(0)=0.5J.

    Thus, the potential energy with respect to zero at x=0x=0 is just
    U(x)=13ax3+0.5J.U(x)=13ax3+0.5J.

    Therefore, the potential energy at x=1mx=1m is
    U(1m)=13(3N/m2)(1m)3+0.5J=1.5J.U(1m)=13(3N/m2)(1m)3+0.5J=1.5J.

Significance In this one-dimensional example, any function we can integrate, independent of path, is conservative. Notice how we applied the definition of potential energy difference to determine the potential energy function with respect to zero at a chosen point. Also notice that the potential energy, as determined in part (b), at x=1mx=1m is U(1m)=1JU(1m)=1J and at x=2mx=2m is U(2m)=8JU(2m)=8J; their difference is the result in part (a).

Check Your Understanding 8.1

In Example 8.1, what are the potential energies of the particle at x=1mx=1m and x=2mx=2m with respect to zero at x=1.5mx=1.5m? Verify that the difference of potential energy is still 7 J.

Systems of Several Particles

In general, a system of interest could consist of several particles. The difference in the potential energy of the system is the negative of the work done by gravitational or elastic forces, which, as we will see in the next section, are conservative forces. The potential energy difference depends only on the initial and final positions of the particles, and on some parameters that characterize the interaction (like mass for gravity or the spring constant for a Hooke’s law force).

It is important to remember that potential energy is a property of the interactions between objects in a chosen system, and not just a property of each object. This is especially true for electric forces, although in the examples of potential energy we consider below, parts of the system are either so big (like Earth, compared to an object on its surface) or so small (like a massless spring), that the changes those parts undergo are negligible if included in the system.

Types of Potential Energy

For each type of interaction present in a system, you can label a corresponding type of potential energy. The total potential energy of the system is the sum of the potential energies of all the types. (This follows from the additive property of the dot product in the expression for the work done.) Let’s look at some specific examples of types of potential energy discussed in Work. First, we consider each of these forces when acting separately, and then when both act together.

Gravitational potential energy near Earth’s surface

The system of interest consists of our planet, Earth, and one or more particles near its surface (or bodies small enough to be considered as particles, compared to Earth). The gravitational force on each particle (or body) is just its weight mg near the surface of Earth, acting vertically down. According to Newton’s third law, each particle exerts a force on Earth of equal magnitude but in the opposite direction. Newton’s second law tells us that the magnitude of the acceleration produced by each of these forces on Earth is mg divided by Earth’s mass. Since the ratio of the mass of any ordinary object to the mass of Earth is vanishingly small, the motion of Earth can be completely neglected. Therefore, we consider this system to be a group of single-particle systems, subject to the uniform gravitational force of Earth.

In Work, the work done on a body by Earth’s uniform gravitational force, near its surface, depended on the mass of the body, the acceleration due to gravity, and the difference in height the body traversed, as given by Equation 7.4. By definition, this work is the negative of the difference in the gravitational potential energy, so that difference is

ΔUgrav=Wgrav,AB=mg(yByA).ΔUgrav=Wgrav,AB=mg(yByA).
(8.4)

You can see from this that the gravitational potential energy function, near Earth’s surface, is

U(y)=mgy+const.U(y)=mgy+const.
(8.5)

You can choose the value of the constant, as described in the discussion of Equation 8.2; however, for solving most problems, the most convenient constant to choose is zero for when y=0,y=0, which is the lowest vertical position in the problem.

Example 8.2

Gravitational Potential Energy of a Hiker The summit of Great Blue Hill in Milton, MA, is 147 m above its base and has an elevation above sea level of 195 m (Figure 8.3). (Its Native American name, Massachusett, was adopted by settlers for naming the Bay Colony and state near its location.) A 75-kg hiker ascends from the base to the summit. What is the gravitational potential energy of the hiker-Earth system with respect to zero gravitational potential energy at base height, when the hiker is (a) at the base of the hill, (b) at the summit, and (c) at sea level, afterward?

Sketch of the profile of Great Blue Hill, Milton, MA. The summit is 195 meters above sea level. The base of the hill is 147 meters below the summit.
Figure 8.3 Sketch of the profile of Great Blue Hill, Milton, MA. The altitudes of the three levels are indicated.

Strategy First, we need to pick an origin for the y-axis and then determine the value of the constant that makes the potential energy zero at the height of the base. Then, we can determine the potential energies from Equation 8.5, based on the relationship between the zero potential energy height and the height at which the hiker is located.

Solution

  1. Let’s choose the origin for the y-axis at base height, where we also want the zero of potential energy to be. This choice makes the constant equal to zero and
    U(base)=U(0)=0.U(base)=U(0)=0.
  2. At the summit, y=147my=147m, so
    U(summit)=U(147m)=mgh=(75×9.8N)(147m)=108kJ.U(summit)=U(147m)=mgh=(75×9.8N)(147m)=108kJ.
  3. At sea level, y=(147195)m=−48my=(147195)m=−48m, so
    U(sea-level)=(75×9.8N)(−48m)=−35.3kJ.U(sea-level)=(75×9.8N)(−48m)=−35.3kJ.

Significance Besides illustrating the use of Equation 8.4 and Equation 8.5, the values of gravitational potential energy we found are reasonable. The gravitational potential energy is higher at the summit than at the base, and lower at sea level than at the base. Gravity does work on you on your way up, too! It does negative work and not quite as much (in magnitude), as your muscles do. But it certainly does work. Similarly, your muscles do work on your way down, as negative work. The numerical values of the potential energies depend on the choice of zero of potential energy, but the physically meaningful differences of potential energy do not. [Note that since Equation 8.2 is a difference, the numerical values do not depend on the origin of coordinates.]

Check Your Understanding 8.2

What are the values of the gravitational potential energy of the hiker at the base, summit, and sea level, with respect to a sea-level zero of potential energy?

Elastic potential energy

In Work, we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position, as given in Equation 7.5. This work involves only the properties of a Hooke’s law interaction and not the properties of real springs and whatever objects are attached to them. Therefore, we can define the difference of elastic potential energy for a spring force as the negative of the work done by the spring force in this equation, before we consider systems that embody this type of force. Thus,

ΔU=WAB=12k(xB2xA2),ΔU=WAB=12k(xB2xA2),
(8.6)

where the object travels from point A to point B. The potential energy function corresponding to this difference is

U(x)=12kx2+const.U(x)=12kx2+const.
(8.7)

If the spring force is the only force acting, it is simplest to take the zero of potential energy at x=0x=0, when the spring is at its unstretched length. Then, the constant is Equation 8.7 is zero. (Other choices may be more convenient if other forces are acting.)

Example 8.3

Spring Potential Energy A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm. (a) How much elastic potential energy does the spring contribute when its length is 23 cm? (b) How much more potential energy does it contribute if its length increases to 26 cm?

Strategy When the spring is at its unstretched length, it contributes nothing to the potential energy of the system, so we can use Equation 8.7 with the constant equal to zero. The value of x is the length minus the unstretched length. When the spring is expanded, the spring’s displacement or difference between its relaxed length and stretched length should be used for the x-value in calculating the potential energy of the spring.

Solution

  1. The displacement of the spring is x=23cm20cm=3cmx=23cm20cm=3cm, so the contributed potential energy is U=12kx2=12(4N/cm)(3cm)2=0.18JU=12kx2=12(4N/cm)(3cm)2=0.18J.
  2. When the spring’s displacement is x=26cm20cm=6cmx=26cm20cm=6cm, the potential energy is U=12kx2=12(4N/cm)(6cm)2=0.72JU=12kx2=12(4N/cm)(6cm)2=0.72J, which is a 0.54-J increase over the amount in part (a).

Significance Calculating the elastic potential energy and potential energy differences from Equation 8.7 involves solving for the potential energies based on the given lengths of the spring. Since U depends on x2x2, the potential energy for a compression (negative x) is the same as for an extension of equal magnitude.

Check Your Understanding 8.3

When the length of the spring in Example 8.3 changes from an initial value of 22.0 cm to a final value, the elastic potential energy it contributes changes by −0.0800J.−0.0800J. Find the final length.

Gravitational and elastic potential energy

A simple system embodying both gravitational and elastic types of potential energy is a one-dimensional, vertical mass-spring system. This consists of a massive particle (or block), hung from one end of a perfectly elastic, massless spring, the other end of which is fixed, as illustrated in Figure 8.4.

A vertical mass spring system is illustrated. The upper end of the spring is attached to the ceiling. A block of mass m is attached to the lower end.  The spring is drawn at two positions. On the left, the mass is in the equilibrium position. To  the right of this, the spring is drawn with the mass pulled down a distance y sub pull. This position of the mass is labeled as h equal to zero. A graph of y as a function of X is shown to the rightly the illustrations, with y equals zero aligned with the equilibrium position in the illustrations. The plot is sinusoidal, with the minimum y at x=0 and even with the lower mass position in the illustrations.
Figure 8.4 A vertical mass-spring system, with the y-axis pointing downward. The mass is initially at an unstretched spring length, point A. Then it is released, expanding past point B to point C, where it comes to a stop.

First, let's consider the potential energy of the system. We need to define the constant in the potential energy function of Equation 8.5. Often, the ground is a suitable choice for when the gravitational potential energy is zero; however, in this case, the highest point or when y=0y=0 is a convenient location for zero gravitational potential energy. Note that this choice is arbitrary, and the problem can be solved correctly even if another choice is picked.

We must also define the elastic potential energy of the system and the corresponding constant, as detailed in Equation 8.7. This is where the spring is unstretched, or at the y=0y=0 position.

If we consider that the total energy of the system is conserved, then the energy at point A equals point C. The block is placed just on the spring so its initial kinetic energy is zero. By the setup of the problem discussed previously, both the gravitational potential energy and elastic potential energy are equal to zero. Therefore, the initial energy of the system is zero. When the block arrives at point C, its kinetic energy is zero. However, it now has both gravitational potential energy and elastic potential energy. Therefore, we can solve for the distance y that the block travels before coming to a stop:

KA+UA=KC+UC 0=0+mgyC+ (12kyC) 2yC =-2mgkKA+UA=KC+UC 0=0+mgyC+ (12kyC) 2yC =-2mgk
A photograph of a bungee jumper.
Figure 8.5 A bungee jumper transforms gravitational potential energy at the start of the jump into elastic potential energy at the bottom of the jump.

Example 8.4

Potential Energy of a Vertical Mass-Spring SystemA block weighing 1.2N1.2N is hung from a spring with a spring constant of 6.0N/m6.0N/m, as shown in Figure 8.4. (a) What is the maximum expansion of the spring, as seen at point C? (b) What is the total potential energy at point B, halfway between A and C? (c) What is the speed of the block at point B?

StrategyIn part (a) we calculate the distance yCyC as discussed in the previous text. Then in part (b), we use half of the y value to calculate the potential energy at point B using equations Equation 8.4 and Equation 8.6. This energy must be equal to the kinetic energy, Equation 7.6, at point B since the initial energy of the system is zero. By calculating the kinetic energy at point B, we can now calculate the speed of the block at point B.

Solution

  1. Since the total energy of the system is zero at point A as discussed previously, the maximum expansion of the spring is calculated to be:
    yC =-2mgk yC =-2(1.2N) (6.0N/m)=-0.40m yC =-2mgk yC =-2(1.2N) (6.0N/m)=-0.40m
  2. The position of yByBis half of the position at yCyC or -0.20m-0.20m. The total potential energy at point B would therefore be:
    UB=mgyB+(12kyB)2 UB=(1.2N)(-0.20m)+ (16(6N/m)(-0.20m)) 2UB =-0.12JUB=mgyB+(12kyB)2 UB=(1.2N)(-0.20m)+ (16(6N/m)(-0.20m)) 2UB =-0.12J
  3. The mass of the block is the weight divided by gravity.
    m=Fwg=1.2N9.8m/s2=0.12kgm=Fwg=1.2N9.8m/s2=0.12kg
    The kinetic energy at point B therefore is 0.12 J because the total energy is zero. Therefore, the speed of the block at point B is equal to
    K=12mv v= 2Km=2(0.12J)(0.12kg)=1.4m/s K=12mv v= 2Km=2(0.12J)(0.12kg)=1.4m/s

SignificanceEven though the potential energy due to gravity is relative to a chosen zero location, the solutions to this problem would be the same if the zero energy points were chosen at different locations.

Check Your Understanding 8.4

Suppose the mass in Equation 8.6 is doubled while keeping the all other conditions the same. Would the maximum expansion of the spring increase, decrease, or remain the same? Would the speed at point B be larger, smaller, or the same compared to the original mass?

Interactive

View this simulation to learn about conservation of energy with a skater! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!

A sample chart of a variety of energies is shown in Table 8.1 to give you an idea about typical energy values associated with certain events. Some of these are calculated using kinetic energy, whereas others are calculated by using quantities found in a form of potential energy that may not have been discussed at this point.

Object/phenomenon Energy in joules
Big Bang 10681068
Annual world energy use 4.0×10204.0×1020
Large fusion bomb (9 megaton) 3.8×10163.8×1016
Hiroshima-size fission bomb (10 kiloton) 4.2×10134.2×1013
1 barrel crude oil 5.9×1095.9×109
1 metric ton TNT 4.2×1094.2×109
1 gallon of gasoline 1.2×1081.2×108
Daily adult food intake (recommended) 1.2×1071.2×107
1000-kg car at 90 km/h 3.1×1053.1×105
Tennis ball at 100 km/h 2222
Mosquito (10−2g at 0.5 m/s)(10−2g at 0.5 m/s) 1.3×10−61.3×10−6
Single electron in a TV tube beam 4.0×10−154.0×10−15
Energy to break one DNA strand 10−1910−19
Table 8.1 Energy of Various Objects and Phenomena
Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
Citation information

© Sep 19, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.