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12/12 do now. An 8. kg ball is fired horizontally from a 2.0 x 10 3 kg cannon initially at rest. After having been fired, the momentum of the ball is 2.40 x 10 3 kg·m/s east (neglect friction.]. Calculate the magnitude of the cannon’s velocity after the ball is fired. [show all work].

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12 12 do now

12/12 do now

  • An 8. kg ball is fired horizontally from a 2.0 x 103 kg cannon initially at rest. After having been fired, the momentum of the ball is 2.40 x 103 kg·m/s east (neglect friction.]. Calculate the magnitude of the cannon’s velocity after the ball is fired. [show all work]


Work energy and power chapter outline

Work, Energy, and Power - Chapter Outline

Lesson 1: Basic Terminology and Concepts

Work

Definition and Mathematics of Work

Calculating the Amount of Work Done by Forces

Potential Energy

Kinetic Energy

Mechanical Energy

Power

Lesson 2: The Work-Energy Theorem

Internal vs. External Forces

The Work-Energy Connection

Analysis of Situations Involving External Forces

Analysis of Situations in Which Mechanical Energy is Conserved

Application and Practice Questions

Bar Chart Illustrations


Objective work

Objective - Work

  • Definition and Mathematics of Work

  • Calculating the Amount of Work Done by Forces

  • Homework – castle learning


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

Definition and Mathematics of Work

force

  • In physics, work is defined as a _________ acting upon an object to ____________ a __________________.

  • In order for a force to qualify as having done work on an object, there must be a displacement and the force must ___________ the displacement

cause

displacement

cause


Work is done

Work is done


Work is not done

Work is not done


Let s practice work or no work

Let’s practice – work or no work

  • A student applies a force to a wall and becomes exhausted.

  • A calculator falls off a table and free falls to the ground.

  • A waiter carries a tray full of beverages above his head by one arm across the room

  • A rocket accelerates through space.

no work

work

no work

work


Calculating the amount of work done by forces

F

θ

d

Calculating the Amount of Work Done by Forces

  • F - is the force in Newton, which causes the displacement of the object.

  • d - is the displacement in meters

  • θ = angle between force and displacement

  • W - is work in N∙m or Joule (J). 1 J = 1 N∙m = 1 kg∙m2/s2

  • Work is a _____________ quantity

  • Work is independent of time the force acts on the object.

W = F∙d∙cosθ

scalar


W f d cos

W = F∙d∙cosθ

Work done – positive, negative or zero work

Positive work

negative work - force acts in the direction opposite the objects motion in order to slow it down.

no work


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

To Do Work, Forces Must Cause Displacements

W = F∙d∙cosθ = 0


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.


The angle in work equation

The angle in work equation

  • It is important to recognize that the angle has a precise definition. It is the angle between the force and the displacement vectors.

F & d are in the same direction,

θ is 0o.

d

F


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

Each path up to the seat top requires the same amount of work. The amount of work done by a force on any object is given by the equation W = F∙d∙cosθ

where F is the force, d is the displacement and θ is the angle between the force and the displacement vector. In all three cases, θ equals to 0o


Example

example

  • A 20.0 N force is used to push a 2.00 kg cart a distance of 5.00 meters. Determine the amount of work done on the cart by the force.

Given: F = 20.0 N; d = 5.00 meters; m = 2.00 kg; θ = 0

Unknown: W = ?

W = F∙dcosθ

W = (20.0 N)(5.00 m)cos0o

W = 100. J

20.0 N


Example1

example

  • How much work is done in lifting a 5.0 kg box from the floor to a height of 1.2 m above the floor?

Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0

Unknown: W = ?

W = F∙dcosθ

F = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 N

W = F∙d = (49 N) (1.2 m) = 59 J


Example2

example

  • A 2.3 kg block rests on a horizontal surface. A constant force of 5.0 N is applied to the block at an angle of 30.o to the horizontal; determine the work done on the block a distance of 2.0 meters along the surface.

  • Given: F = 5.0 N;

    m = 2.3 kg

    d = 2.0 m

    θ= 30o

5.0 N

30o

2.3 kg

  • unknown:

    • W = ? J

  • Solve:

  • W = F∙d∙cosθ

  • W = (5.0 N)(2.0 m)(cos30o) = 8.7 J


Practice

practice

  • Matt pulls block along a horizontal surface at constant velocity. The diagram show the components of the force exerted on the block by Matt. Determine how much work is done against friction.

  • Given: Fx = 8.0 N

    Fy = 6.0 N

    dx = 3.0 m

6.0 N

  • unknown: W = ? J

W = Fxdx

W = (8.0 N)(3.0 m) = 24 J

8.0 N

3.0 m


Example3

example

  • A neighbor pushes a lawnmower four times as far as you do but exert only half the force, which one of you does more work and by how much?

Wyou = Fd

Wneighbor = (½ F)(4d) = 2 Fd = 2 Wyou

The neighbor, twice as much


Force vs displacement graph

Force vs. displacement graph

  • The area under a force versus displacement graph is the work done by the force.

Example: a block is pulled along a table with 10. N over a distance of 1.0 m.

W = Fd = (10. N)(1.0 m) = 10. J

work

Force (N)

Displacement (m)

height

base

area


Class work

Class work

  • Work practice


12 13 do now

12/13 do now

  • Work, energy, and power packet –

    • page 1-4


Objectives

objectives

  • Describe potential energy

    • gravitational potential energy

    • Hooke’s Law

    • Elastic potential energy

  • Homework – castle learning

  • Reminder: the momentum and collisions packet was due last Friday. 80% credit if hand in today. Last day of collection - Friday 12/16 for 50% credit

  • Food drive ends 12/16 – every two items for one extra credit on your average


Potential energy

Potential energy

  • An object can store energy as the result of its position. ________________________ is the stored energy of position possessed by an object.

  • Two form:

    • Gravitational

    • Elastic

Potential energy


Gravitational potential energy

Gravitational potential energy

vertical position (height).

  • Gravitational potential energy is the energy stored in an object as the result of its _________________________

  • The energy is stored as the result of the _____________ attraction of the Earth for the object.

  • The work done in raising an object must result in an increase in the object's _______________________

  • The gravitational potential energy of the box is dependent on two variables:

    • The mass of the object

    • The height of the object

  • Equation: ______________________

    • m: mass, in kilograms

    • h: height, in meters

    • g: acceleration of gravity = 9.81 m/s2

gravitational

gravitational potential energy

PEgrav = m∙g∙h


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

GPE

  • GPE = mgh

  • The equation shows that . . .

  • . . . the more gravitational potential energy it’s got.

  • the more mass a body has

  • or the stronger the gravitational field it’s in

  • or the higher up it is


Unit of energy

Unit of energy

  • The unit of energy is the same as work: _______

  • 1 joule = 1 (kg)∙(m/s2)∙(m) = 1 Newton ∙ meter

  • 1 joule = 1 (kg)∙(m2/s2)

Joules

Work and energy has the same unit


Gravitational potential energy is relative

Gravitational potential energy is relative

  • To determine the gravitational potential energy of an object, a _______ height position must first be assigned.

  • Typically, the ___________ is considered to be a position of zero height.

  • But, it doesn’t have to be:

    • It could be relative to the height above the lab table.

    • It could be relative to the bottom of a mountain

    • It could be the lowest position on a roller coaster

zero

ground


Example4

example

  • How much potential energy is gained by an object with a mass of 2.00 kg that is lifted from the floor to the top of 0.92 m high table?

  • Known:

  • m = 2.00 kg

  • h = 0.92 m

  • g = 9.81 m/s2

Solve:

∆PE = mg∆h

∆PE = (2.00 kg)(9.81m/s2)(0.92 m) = 18 J

  • unknown:

  • PE = ? J


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

  • The graph of gravitational potential energy vs. vertical height for an object near Earth's surface gives the weight of the object.

The weight of the object is the slope of the line.

Weight = 25 J/1.0 m = 25 N

m = weight / g = 2.5 kg


Elastic potential energy

Elastic potential energy

  • Elastic potential energy is the energy stored in ______________ materials as the result of their stretching or compressing.

  • Elastic potential energy can be stored in

    • Rubber bands

    • Bungee cores

    • Springs

    • trampolines

elastic


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

  • Springs are a special instance of a device that can store elastic potential energy due to either compression or stretching.

  • A force is required to compress or stretch a spring; the more compression/stretch there is, the more force that is required to compress it further.

  • For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k).


Hooke s law

Hooke’s Law

F = kx

Spring force = spring constant x displacement

  • F in the force needed to displace (by stretching or compressing) a spring x meters from the equilibrium (relaxed) position. The SI unit of F is Newton.

  • k is spring constant. It is a measure of stiffness of the spring. The greater value of k means a stiffer spring because more force is needed to stretch or compress it that spring. The Si units of k are N/m. depends on the material made up of the spring. k is in N/m

  • x the distance difference between the length of stretched/compressed spring and its relaxed (equilibrium) spring.


Example5

example

  • Determine the x in F = kx


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

force

elongation

F = kx

  • Spring force is directly proportional to the elongation of the spring (displacement)

The slope represents spring constant: k = F / x


Caution

elongation

force

caution

  • Sometimes, we might see a graph such as this:

The slope represents the inverse of spring constant:

Slope = 1/k = x / F


Example6

example

  • Given the following data table and corresponding graph, calculate the spring constant of this spring.


Example7

example

  • A 20.-newton weight is attached to a spring, causing it to stretch, as shown in the diagram. What is the spring constant of this spring?


12 14 do now

12/14 do now

  • Which of the following statements are true about work? Include all that apply.

  • Work is a form of energy.

  • Units of work would be equivalent to a Newton times a meter.

  • A kg•m2/s2 would be a unit of work.

  • Work is a time-based quantity; it is dependent upon how fast a force displaces an object.

  • Superman applies a force on a truck to prevent it from moving down a hill. This is an example of work being done.

  • An upward force is applied to a bucket as it is carried 20 m across the yard. This is an example of work being done.

  • A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop of the Shockwave ride. This is an example of work being done.


Objectives1

objectives

  • Hooke’s law

  • Potential energy

  • Kinetic energy

  • Lab – determine spring constant using Hooke’s law

  • Homework – castle leaning

  • Reminder: the momentum and collisions packet was due last Friday. 70% credit if hand in today. Last day of collection - Friday 12/16 for 50% credit

  • Food drive ends 12/16 – every two items for one extra credit on your average

  • Quiz on Friday – work, potential and kinetic energy, Hooke’s Law – questions are from castle learning assignments


Example8

example

  • A spring has a spring constant of 25 N/m.  What is the minimum force required to stretch the spring 0.25 meter from its equilibrium position?


Example9

example

  • The graph below shows elongation as a function of the applied force for two springs, A and B. Compared to the spring constant for spring A, the spring constant for spring B is

  • smaller

  • larger

  • the same


Elastic potential energy in a spring

Elastic potential energy in a spring

  • Elastic potential energy is the Work done on the spring.

    PEs = Favg∙d = Favg∙x = (½ k∙x)∙x = ½ kx2

    Note: F is the average force

    • k: spring constant

    • x: amount of compression or extension relative to equilibrium position

PEs = ½ k∙x2


Elastic potential energy is directly proportional to x 2

Elastic potential energy

elongation

Elastic potential energy is directly proportional to x2


Example10

example

  • A spring has a spring constant of 120 N/m.  How much potential energy is stored in the spring as it is stretched 0.20 meter?

Given:

k = 120 N/m

x = 0.20 m

Unknown:

PEs = ? J

Solve:

PEs = ½ kx2

PEs = ½ (120 N/m)(0.20 m)2

PEs = 2.4 J


Example11

example

  • The unstretched spring in the diagram has a length of 0.40 meter and a spring constant k.  A weight is hung from the spring, causing it to stretch to a length of 0.60 meter.  In terms of k, how many joules of elastic potential energy are stored in this stretched spring?

PEs = ½ kx2

PEs = ½ k(0.20 m)2

PEs = (0.020 k) J


Example12

example

  • Determine the potential energy stored in the spring with a spring constant of 25.0 N/m when a force of 2.50 N is applied to it.

Solve:

PEs = ½ kx2

To find x, use Fs = kx,

(2.50 N) = (25.0 N/m)(x)

x = 0.100 m

PEs = ½ (25.0 N/m)(0.100 m)2

PEs = 0.125 J

Given:

Fs = 2.50 N

k = 25.0 N/m

Unknown:

PEs = ? J


Example13

example

  • As shown in the diagram, a 0.50-meter-long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight. If 15 joules of energy are stored in the stretched spring, what is the value of the spring constant?

PE = ½ kx2

15 J = ½ k (0.50 m)2

k = 120 N/m


Example14

example

  • A 10.-newton force is required to hold a stretched spring 0.20 meter from its rest position. What is the potential energy stored in the stretched spring?

Given:

F = 10. N

x = 0.20 m

Unknown: PEs = ? J

PEs = ½ kx2 = ½ (kx)(x)

PEs = ½ Fx

PEs = ½ (10. N)(.20 m)

PEs = 1.0 J


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

  • A force of 0.2 N is needed to compress a spring a distance of 0.02 meter. What is the potential energy stored in this compressed spring?

Given:

F = 0.2 N

x = 0.02 m

Unknown: PEs = ? J

PEs = ½ kx2 = ½ (kx)(x)

PEs = ½ Fx

PEs = ½ (0.2 N)(0.02 m)

PEs = 0.002 J


Kinetic energy

KE = ½ mv2

Kinetic energy

motion

  • Kinetic energy is the energy of _______.

  • An object which has motion - whether it be vertical or horizontal motion - has kinetic energy.

  • The equation for kinetic energy is:

    __________________

    • Where KE is kinetic energy, in joules

    • v is the speed of the object, in m/s

    • m is the mass of the object, in kg


Kinetic energy1

Kinetic Energy

  • KE = ½ mv2

  • The equation shows that . . .

  • . . . the more kinetic energy it has.

  • the more mass a body has

  • or the faster it’s moving


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

Kinetic energy

speed

  • KE is proportional to v2, so doubling the speed quadruples kinetic energy, and tripling the speed makes it nine times greater.


Example15

Example

  • A 55 kg toy sailboat is cruising at 3 m/s. What is its kinetic energy?

    Note: Kinetic energy (along with every other type of energy) is a scalar, not a vector!

KE= ½ mv2

KE = 0.5 (55) (3) 2 = 247.5 J


Example16

example

  • How much kinetic energy is possessed by a 2.7 kg cart traveling at 1.5 m/s?

Known:

  • m = 2.7 kg

  • v = 1.5 m/s

    Unknown:

  • KE = ? J

Solve:

KE = ½ mv2

KE = ½ (2.7 kg)(1.5 m/s)2

KE = 3.0 J


Example17

example

  • How much kinetic energy is possessed by a 27 N cart traveling at 1.5 m/s?

Solve:

KE = ½ mv2

To find mass, we can use

Weight = mg

m = weight / g = 27 N / 9.81 m/s2

m = 2.8 kg

KE = ½ (2.8 kg)(1.5 m/s)2

KE = 3.1 J

Known:

  • weight = 27 N

  • v = 1.5 m/s

    Unknown:

  • KE = ? J


Example18

example

  • A cart of mass m traveling at a speed v has kinetic energy KE.  If the mass of the cart is doubled and its speed is halved, the kinetic energy of the cart will be

  • half as great

  • twice as great

  • one-fourth as great

  • four times as great


Example19

example

  • An object moving at a constant speed of 25 meters per second possesses 450 joules of kinetic energy. What is the object's mass?

Known:

  • KE = 450 J

  • v = 25 m/s

    Unknown:

  • m = ? kg

Solve:

KE = ½ mv2

450 J = ½ (m)(25 m/s)2

m = 1.4 kg


Example20

example

  • Which graph best represents the relationship between the kinetic energy, KE, and the velocity of an object accelerating in a straight line?

a

b

c

d


12 15 do now

force

weight

elongation

mass

12/15 do now

  • Determine the meaning of slope in each graph

elongation

A

B

force

Gravitational potential energy

C

D

height


Objectives2

objectives

  • Mechanical energy

  • Energy practice

  • Power

  • practice

  • Lab – determine spring constant using Hooke’s law

  • Homework – castle leaning

  • Reminder: the momentum and collisions packet was due last Friday. 60% credit if hand in today. Last day of collection - Friday 12/16 for 50% credit

  • Food drive ends tomorrow – every two items for one extra credit on your average

  • Quiz on Friday – work, potential and kinetic energy, Hooke’s Law – questions are from castle learning assignments

  • Last day to take/retake the momentum impulse test is Thursday 12/22. You must make test corrections and do the castle learning review before you retake the test


Mechanical energy

Mechanical Energy

  • Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position) or both


Mechanical energy as the ability to do work

Mechanical Energy as the Ability to Do Work

  • Any object that possesses mechanical energy - whether it is in the form of potential energy or kinetic energy - is able to do work.


The total mechanical energy

The Total Mechanical Energy

  • The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

TME = PE + KE

TME = PEgrav + PEspring + KE


An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

  • The diagram shows the motion of Brie as she glides down the hill and makes one of her record-setting jumps.

TME =

TME =

TME =

TME =

TME =


Lab 14 hooke s law 1

Lab 14 – Hooke’s Law (1)

  • Purpose: To determine the spring constant of a given spring.

  • Material: spring, masses, meter stick.

  • Procedure: Hook different masses on the spring, record the force Fs (mg) and corresponding elongation x. Plot the graph of Force vs. elongation

  • Data section: should contain colomns: force applied, elongation.

    • Data measured directly from the experiment. The units of measurements in a data table should be specified in column heading only.

  • Data analysis: Graph force vs. elongation on graph paper, answer following questions:

    • What does the slope mean in Force vs. elongation graph?

    • Determine the spring constant


  • An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    Force vs. elongation

    Force (N)

    Elongation (m)


    Mechanical energy1

    Mechanical Energy

    • Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position) or both


    Mechanical energy as the ability to do work1

    Mechanical Energy as the Ability to Do Work

    • Any object that possesses mechanical energy - whether it is in the form of potential energy or kinetic energy - is able to do work.


    The total mechanical energy1

    The Total Mechanical Energy

    • The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

    TME = PE + KE

    TME = PEgrav + PEspring + KE


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • The diagram shows the motion of Brie as she glides down the hill and makes one of her record-setting jumps.

    TME =

    TME =

    TME =

    TME =

    TME =


    Class work1

    Class work

    • Work, energy, and power packet – page 7-8


    Power

    Power

    • Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

    • The standard metric unit of power is the Watt.


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • All machines are typically described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects.

    • The power rating of a car relates to how rapidly the car can accelerate the car.

    • Some people are more power-full than others. That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time


    Example21

    example

    • Ben Pumpiniron elevates his 80-kg body up the 2.0-meter stairwell in 1.8 seconds. What is his power?

    It can be assumed that Ben must apply an (80 kg x 9.81 m/s2) -Newton downward force upon the stairs to elevate his body.


    Another equation for power

    Another equation for power


    Example22

    example

    • Two physics students, Will N. Andable and Ben Pumpiniron, are in the weightlifting room. Will lifts the 100-pound barbell over his head 10 times in one minute; Ben lifts the 100-pound barbell over his head 10 times in 10 seconds. Which student does the most work? ______________ Which student delivers the most power? ______________ Explain your answers.


    Example23

    example

    • When doing a chin-up, a physics student lifts her 42.0-kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps?


    Kilowatt hour is unit for energy

    kilowatt-hour is unit for energy

    • Your household's monthly electric bill is often expressed in kilowatt-hours. One kilowatt-hour is the amount of energy delivered by the flow of l kilowatt of electricity for one hour. Use conversion factors to show how many joules of energy you get when you buy 1 kilowatt-hour of electricity.


    What is energy

    What is energy?

    ability

    • Energy is the __________to do work.

    • Energy is a _________quantity.

    • When work is done on or by a system, the total energy of the system is changed.

    scalar


    Class work2

    Class work


    Lesson 2 the work energy theorem

    Lesson 2: The Work-Energy Theorem

    • Kinetic energy - Worktheorem

    • conservative vs. non-conservative Forces

    • The Work-Energy Connection

      • Analysis of Situations Involving non-conservative Forces

      • Analysis of Situations Involving conservative forces

      • Application and Practice Questions


    Kinetic energy work theorem

    Kinetic energy - Work theorem

    • The net work done on an object equals to the change in the object’s kinetic energy

    Wnet = ∆KE = KE2 – KE1

    Fd = KE2 – KE1


    Practice problem 1

    Practice Problem #1

    • A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car. 

    Wnet = ∆KE = KEf - KEi

    (-8000N) • d = -312 500 0 J

    d = 39.1 m


    Practice problem 2

    Practice Problem #2

    • At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount.

    Wnet = ∆KE = KEf - KEi


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • The above problems have one thing in common: there is a force which does work over a distance in order to remove mechanical energy from an object.

    • The force acts opposite the object's motion and thus does negative work which results in a loss of the object's total amount of mechanical energy. In each situation, the work is related to the kinetic energy change.

    TMEi + Wext = TMEf

    KEi + Wext = 0 J

    ½ •m•vi2 + F•d•cos(180o) = 0 J

    F•d = ½ •m•vi2

    d vi2

    Stopping distance is dependent upon the square of the velocity.


    Practice1

    practice

    4 m x 22 = 16 m

    4 m x 32 = 36 m

    4 m x 42 = 64 m

    4 m x 52 = 100 m


    Conservative vs non conservative forces

    conservative vs. non-conservative Forces

    • There are a variety of ways to categorize all the types of forces.

    • Contact force: Forces that arise from the physical contact of two objects.

    • Field force exist between objects, even in the absence of physical contact between the objects.

    • In this lesson, we will learn how to categorize forces based upon whether or not their presence is capable of changing an object's total mechanical energy.

    • Conservative force can never change the total mechanical energy of an object

    • Non-conservative forces will change the total mechanical energy of the object


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • For our purposes, we will simply say that non-conservative forces include the applied force, normal force, tension force, friction force, and air resistance force. And for our purposes, the conservative forces include the gravity forces, magnetic force, electrical force, and spring force.


    The big concept

    The big concept

    • When net work is done upon an object by an non-conservative force, the total mechanical energy (KE + PE) of that object is changed.

      • If the work is positive work, then the object will gain energy.

      • If the work is negative work, then the object will lose energy.

      • The gain or loss in energy can be in the form of potential energy, kinetic energy, or both.

      • The work done will be equal to the change in mechanical energy of the object.


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • When the only type of force doing net work upon an object is conservative force (for example, gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant. In such cases, the object's energy changes form.

    • For example, as an object is "forced" from a high elevation to a lower elevation by gravity, some of the potential energy of that object is transformed into kinetic energy. Yet, the sum of the kinetic and potential energies remain constant. This is referred to as energy conservation.


    Situations involving non conservative forces

    Situations Involving non-conservative Forces

    • The quantitative relationship between work and mechanical energy is expressed by the following equation:

    TMEi + Wext = TMEf

    or

    KEi + PEi + Wext = KEf + PEf

    The equation states that the initial amount of total mechanical energy (TMEi) plus the work done by external forces (Wext) is equal to the final amount of total mechanical energy (TMEf). Note the work can be positive and negative.


    Example24

    example


    Example25

    example


    Example26

    example


    Example27

    example


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • In each of these examples, a non-conservative force does work upon an object over a given distance to change the total mechanical energy of the object.

    • If the work is positive, then the object gains mechanical energy. The amount of energy gained is equal to the work done on the object.

    • If the work is negative, then the object loses mechanical energy. The amount of mechanical energy lost is equal to the work done on the object.

    • In general, the total mechanical energy of the object in the initial state (prior to the work being done) plus the work doneequals the total mechanical energy in the final state.


    Example28

    example

    • A block weighing 15 N is pulled to the top of an incline that is 0.20 meter above the ground, as shown below. If 5.0 joules of work are needed to pull the block the full length of the incline, how much work is done against friction?

    KEi + PEi + Wext = KEf + PEf

    The external forces are applied force and friction force

    0 + 0 + Wapp + Wf = 0 + (15 N)(0.20 m)

    5.0 J + Wf = 3.0J

    Wf = -2.0 J

    2.0 J of work is done to overcome friction


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    example

    • A shopping cart full of groceries is sitting at the top of a 2.0-m hill. The cart begins to roll until it hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart and hits a parked car with an average force of 500 N. How deep a dent is made in the car (i.e., over what distance does the 500 N force act upon the can of peaches before bringing it to a stop)?

    Initially:PE = (0.25 kg)(9.81 m/s2)(2 m) = 4.9 J

    KE = 0 J (the peach can is at rest)

    Finally: PE = 0 J (the can's height is zero)

    KE = 0 J (the peach can is at rest)

    Wext = (500 N) • (d) • cos 180 = -(500 N)•d

    KEi + PEi + Wext = KEf + PEf4.9 J + (-500 N)∙d = 0 J

    d = 0.0098 m


    Example29

    Example

    In the diagram below, 450. joules of work is done raising a 72-newton weight a vertical distance of 5.0 meters. How much work is done to overcome friction as the weight is raised?

    KEi + PEi + Wext = KEf + PEf

    There are two external forces: applied force and friction force

    The applied force did 450 J of work:

    0 + 0 + 450 J + Wf = 0 + (72 N)(5.0m)

    450 J + Wf = (72 N)(5.0 m)

    Wf = -90 J

    90 J of work is done to overcome friction


    Brief review

    Brief review

    • When external force does work on an object, its total mechanical energy is changed.

    • Note: the mechanical energy can be either potential energy (in which case it could be due to springs or gravity) or kinetic energy. Given this fact, the above equation can be rewritten as

    • KEi + PEi + Wext = KEf + PEf

    • Note: the work done by external forces can be a positive or a negative work term. it is dependent upon the angle between the force and the motion. If the angle is 180o as it occasionally is, then the work term will be negative. If the angle is 0o, then the work term will be positive.


    Example external force

    B

    0.50 m

    Example – external force

    • A box with a mass of 0.04 kg starts from rest at point A and travels 5.00 meters along a uniform track until coming to rest at point B, as shown in the picture. Determine the magnitude of the frictional force acting on the box. (assume the frictional force is constant.)

    • Given:

      • hA = 0.80 m

      • hB= 0.50 m

      • d = 5.00 m

      • m=0.04 kg

    • Unknown:

    • Ff = ? N

    KEi + PEi + Wext = KEf + PEf

    0+mg(0.80m)+Wf=0+ mg(0.50m)

    Wf= -0.12 J

    0.12 J of work is done to overcome friction

    Wf = Ff∙d = 0.12 J

    Ff∙(5.00m) = 0.12 J

    Ff = 2.4 x 10-2 N

    A

    0.80 m


    Example30

    example

    • A block weighing 40. newtons is released from rest on an incline 8.0 meters above the horizontal, as shown in the diagram below. If 50. joules of heat is generated as the block slides down the incline, what is the maximum kinetic energy of the block at the bottom of the incline?

    KEi + PEi + Wext = KEf + PEf

    The external force is friction force only

    0 + (40.N)(8.0m) + Wf = KEf + 0

    320 J – 50 J = KEf

    KEf = 270 J


    Example31

    example

    • A block weighing 15 newtons is pulled to the top of an incline that is 0.20 meter above the ground, as shown below. If 4.0 joules of work are needed to pull the block the full length of the incline, how much work is done against friction?

    KEi + PEi + Wext = KEf + PEf

    The external forces are applied force and friction force

    0 + 0 + Wapp + Wf = 0 + (15 N)(0.20 m)

    4.0 J + Wf = 3.0J

    Wf = -1.0 J

    1.0 J of work is done to overcome friction


    Lesson 2 the work energy theorem1

    Lesson 2: The Work-Energy Theorem

    • The Work-Energy Connection

      • Analysis of Situations in Which Mechanical Energy is Conserved

      • Application and Practice Questions


    Analysis of situations in which mechanical energy is conserved

    Analysis of Situations in Which Mechanical Energy is Conserved

    • Whenever work is done upon an object by an external force (or nonconservative force), there will be a change in the total mechanical energy of the object.

    KEi + PEi + Wext = KEf + PEf

    If onlyinternal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved.

    Wext = 0

    KEi + PEi = KEf + PEf


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    KEi + PEi = KEf + PE

    KEi + PEi = KEf + PEf

    • In these situations, the sum of the kinetic and potential energy is everywhere the same.

    KEi - KEf = PEf - PEi

    Decrease in KE = Increase in PE

    • As the potential energy is increased, the kinetic energy is decreased.

    • As the potential energy is decreased, the kinetic energy is increased.

    • The increase in potential energy equals to the decrease in kinetic energy and vise versa.


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • We would say that energy is transformed or changes its form from kinetic energy to potential energy (or vice versa); yet the total amount present is conserved - i.e., always the same.


    The example of pendulum motion

    The Example of Pendulum Motion

    • Consider a pendulum bob swinging to and fro on the end of a string. There are only two forces acting upon the pendulum bob. Gravity (an internal force) acts downward and the tensional force (an external force) pulls upwards towards the pivot point. The external force does not do work since at all times it is directed at a 90-degree angle to the motion.


    The pendulum

    The pendulum

    mechanical energy

    • The sum of the kinetic and potential energies in system is called the total ______________________________.

    • In the case of a pendulum, the total mechanical energy (KE + PE) is _________________: at the highest point, all the energy is potential energy, at the lowest point, all the energy is kinetic energy.

    constant


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • As the 2.0-kg pendulum bob in the above diagram swings to and fro, its height and speed change. Use energy equations and the above data to determine the blanks in the above diagram.

    0.153

    0

    0.306

    0.306

    1.73

    2.45


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    use the heights and the speeds given in the table below to fill in the remaining cells at the various locations in a 0.200-kg bob's trajectory.

    3.920

    0

    3.920

    3.162

    2.920

    1.000

    4.134

    2.211

    1.709

    3.920

    1.748

    2.172

    3.920

    4.660

    5.675

    0.700

    3.220

    3.920


    Example32

    Example

    • As the pendulum swings from position A to position C as shown in the diagram, what is the relationship of kinetic energy to potential energy? [Neglect friction.]

    • The kinetic energy decreases more than the potential energy increases.

    • The kinetic energy increases more than the potential energy decreases.

    • The kinetic energy decrease is equal to the potential energy increase.

    • The kinetic energy increase is equal to the potential energy decrease.


    Example33

    example

    • A pendulum is pulled to the side and released from rest. Sketch a graph best represents the relationship between the gravitational potential energy of the pendulum and its displacement from its point of release.

    PE

    pos


    Example34

    Example

    • In the diagram, an ideal pendulum released from point A swings freely through point B. Compared to the pendulum's kinetic energy at A, its potential energy at B is

    • half as great

    • twice as great

    • the same

    • four times as great


    Roller coaster

    Roller coaster

    • A roller coaster operates on the principle of energy transformation. Work is initially done on a roller coaster car to lift the car to the first and highest hill. The roller coaster car has a large quantity of potential energy and virtually no kinetic energy as it begins the trip down the first hill. As the car descents hills and loops, it potential energy is transformed into kinetic energy; as the car ascends hills and loops, its kinetic energy is transformed into potential energy. The total mechanical energy of the car is _______________ when friction is ignored.

    conserved


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • Conservation of energy on a roller coaster ride means that the total amount of mechanical energy is the same at every location along the track.


    An 8 kg ball is fired horizontally from a 2 0 x 103 kg cannon initially at rest after having been fired the momentum

    • The total mechanical energy of the roller coaster car is a constant value of 40 000 Joules.


    The skier

    The skier

    • Transformation of energy from the potential to the kinetic also occurs for a ski jumper. As a ski jumper glides down the hill towards the jump ramp and off the jump ramp towards the ground, potential energy is transformed into kinetic energy. If friction can be ignored, the total mechanical energy is ______________________.

    conserved.


    A free falling object

    A free falling object

    • If a stationary object having mass m is located a vertical distance h above Earth’s surface, the object has initial PE = mgh and KE = 0. as object falls, its PE ___________ and KE ________________. The total mechanical energy is conserved.

    increases

    decreases


    Energy conversion of a free falling object

    Energy conversion of a free falling object

    The graph shows as a ball is dropped, how its energy is transformed.

    • The total mechanical energy remains _____________.

    • GPE decreases as KE increases

    constant


    Example35

    example

    • A 3.0-kilogram object is placed on a frictionless track at point A and released from rest. (Assume the gravitational potential energy of the system to be zero at point C.) Calculate the kinetic energy of the object at point B.

    KEi + PEi = KEf + PEf

    0 + mg(3.0m) = KEf + mg(1.0m)

    KEf = mg∆h = (3.0 kg)(9.81 m/s2)(3.0 m – 1.0 m) = 59 J

    (KE gained at B is potential lost from A to B


    Example36

    example

    • A 250.-kilogram car is initially at rest at point A on a roller coaster track. The car carries a 75-kilogram passenger and is 20. meters above the ground at point A. [Neglect friction.] Compare the total mechanical energy of the car and passenger at points A, B, and C.

    • The total mechanical energy is less at point C than it is at points A or B.

    • The total mechanical energy is greatest at point A.

    • The total mechanical energy is the same at all three points.

    • The total mechanical energy is greatest at point B.


    Example37

    example

    • The diagram represents a 0.20-kilogram sphere moving to the right along a section of a frictionless surface. The speed of the sphere at point A is 3.0 meters per second.

    • Approximately how much kinetic energy does the sphere gain as it goes from point A to point B?

    KEi + PEi = KEf + PEf

    KEf - KEi = PEi - PEf

    The KE gained is PE lost

    ∆KE = ∆PE = mg∆h

    KE = (0.20 kg)(9.81 m/s2)(1.00 m) = 2.0 J


    Example38

    example

    • A 1.0 kg mass falls freely for 20. meters near the surface of Earth. What is the total KE gained by the object during its free fall?

    KEi + PEi = KEf + PEf

    KEf - KEi = PEi - PEf

    ∆KE (gained) = GPE (lost)

    ∆KE = (1.0 kg)(9.81 m/s2)(20.m-0) = 2.0x102 J


    Example39

    example

    • Base your answer to the question on the information and diagram. A 250.-kilogram car is initially at rest at point A on a roller coaster track. The car carries a 75-kilogram passenger and is 20. meters above the ground at point A. [Neglect friction.] Calculate the speed of the car and passenger at point B.

    20. m/s


    Example40

    example

    • The diagram shows points A, B, and C at or near Earth’s surface. As a mass is moved from A to B, 100. joules of work are done against gravity. What is the amount of work done against gravity as an identical mass is moved from A to C?

    KEi + PEi + Wext = KEf + PEf

    Wext = KEf + PEf – (KEi + PEi)

    Wext = 100 J - 0

    Wext = 100 J


    Example41

    example

    • A 20.-kilogram object strikes the ground with 1960 joules of kinetic energy after falling freely from rest.  How far above the ground was the object when it was released?

    10. m


    Example42

    Example

    • A 55.0-kilogram diver falls freely from a diving platform that is 3.00 meters above the surface of the water in a pool. When she is 1.00 meter above the water, what are her gravitational potential energy and kinetic energy with respect to the water's surface?

    In this situation, the force doing the work is gravity, which is an internal force.

    KEi + PEi = KEf + PEf

    KEi = 0J;

    PEi = mgh=(55.0kg)(9.81m/s2)(3.00m) = 1619 J

    PEf = mgh = (55.0kg)(9.81m/s2)(1.00m) = 540. J

    KEf = KEi + PEi - PEf = 1080 J


    When work is done by elastic force energy is conserved

    When work is done by elastic force, energy is conserved

    • A person does 100 joules of work in pulling back the string of a bow. What will be the initial speed of a 0.5-kilogram arrow when it is fired from the bow?

    KEi + PEi + Wext = KEf + PEf

    0 + 0 + 100 J = KEf + 0

    KEf = 100 J

    KE = ½ mv2

    100 J = ½ (0.5 kg)v2 v = 20 m/s


    Example43

    example

    • The diagram shows a 0.1-kilogram apple attached to a branch of a tree 2 meters above a spring on the ground below. The apple falls and hits the spring, compressing it 0.1 meter from its rest position. If all of the gravitational potential energy of the apple on the tree is transferred to the spring when it is compressed, what is the spring constant of this spring?

    KEi + PEi = KEf + PEf

    0 + mgh = 0 + ½ kx2

    mgh = ½ kx2

    (0.1 kg)(9.81m/s2)(2m) = ½∙k (0.1m)2

    k = 400 N/m


    Example44

    example

    • A spring in a toy car is compressed a distance, x. When released, the spring returns to its original length, transferring its energy to the car. Consequently, the car having mass m moves with speed v. Derive the spring constant, k, of the car’s spring in terms of m, x, and v. [Assume an ideal mechanical system with no loss of energy.]

    KEi + PEi = KEf + PEf

    0 + ½ kx2 = ½ mv2 + 0

    ½ kx2 = ½ mv2

    k = mv2/x2


    Example45

    example

    • A person does 64 joules of work in pulling back the string of a bow. What will be the initial speed of a 0.5-kilogram arrow when it is fired from the bow?

    16 m/s


    Energy review

    energy review

    • Work = Fdcosθ (θ is the angle between F & d)

    • Power = W/t = Fv

    • PEg = mgh

    • Fs = kx

    • PEs = ½ kx2

    • KE = ½ mv2

    • When there is no external force:

      • Mechanical energy is conserved: TMEi = TMEf

      • KEi + PEi = KEf + PEf

    • When there is external force (friction):

      • Mechanical energy is not conserved

      • Ei + W(applied force & friction force) = Ef


    12 15 do now1

    12/15 do now

    • Which pair of quantities can be expressed using the same units?

    • work and kinetic energy

    • power and momentum

    • impulse and potential energy

    • acceleration and weight


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