Chapters 10-11

Work and Energy

• Work and Kinetic Energy

• Work of Constant Forces

• Kinetic Energy

• Work-Energy Theorem

• Elastic Collisions

• Work of Varying Forces:• Work of Varying Forces:

• Hooke’s Law

• Potential Energy

• Gravitational Potential Energy

• Elastic Potential Energy

• Conservative and Nonconservative Forces

• Force and Potential Energy. Energy Diagrams

• Power

Why an alternative treatment based on Energy and Momentum?

• In this lecture set, we’ll refer to the same physical situations as in the previously, but

from a different perspective: using energy

• Energy and momentum are physical quantities that obey conservation laws such

that they can be used to describe how the motion is transmitted between interacting

objects while overall they stay constant

• Conservation laws are still valid even when classical mechanics fails

Ex: a) When an object moves along a rough surface its energy is transmitted to the ground

so, if there is no other source to replenish its energy, the object will eventually stop: the

total energy in the system object-surface is conserved. b) A freely falling object interacts total energy in the system object-surface is conserved. b) A freely falling object interacts

with the Earth which transfers energy to the object such that the object will move faster

and faster: the total energy in the system object-Earth is conserved.

• In the previous lectures, we defined conservation of momentum. In this lecture we’ll

study conservation of energy:

If a system is isolated, its total energy – that is, the sum of all its energies present

in various forms – remains the same, albeit the form of energies may change

• In mechanics we are particularly interested in the energy associated with motion –

Kinetic Energy – and how it changes – via the Work done on the system by forces.

Work – Constant force. 1D motion

• A force acting on an object may intermediate an exchange of energy measured by the

work done by the force: a scalar quantity with dimension of energy

• The work done by a force depends on the force, on the displacement of the object,

and on whether the force contributes to the respective displacement

• The motion of an object is characterized by a unique vector displacement, but it can

be acted by a multitude of forces, so each force will do work along that displacement

• Let’s start with the simplest situation:

Def: If a constant force F acts on an object moving in a straight line a displacement

∆r, with the force making an angle θ with the direction of motion, the work done by ∆r, with the force making an angle θ with the direction of motion, the work done by

the force on the object is given by the dot product

cosW F r F r θ= ⋅∆ = ∆�

�

1 Joule (J) N mSI

W = = ⋅

F�

mθ Fcosθ F

�

• Notice that, by this definition, the work is only the work done by the component of

the force along the direction of motion

r∆�

r∆�

• The work describes an exchange of energy in which one object in the force-action

pair looses energy and the other gains energy

• The nature of the work is given by its sign. Thus, work can be negative (F decreases

the energy of the object), zero (F has no effect on the energy of the object), or positive

(F increases the energy of the object)

• The sign of the work is given by cosθ in the definition:

F�

v�

F�

v�

F�

v�

θ < 90° → WF >0 θ = 90° → WF = 0 θ > 90° → WF <0

Work – Comments

v v v

force helps the motion no effect on motion force against the motion

Ex: As long as this

person does not lift or

lower the bag of

groceries, he is not

doing work on it, since

the force FP he exerts

has no component in the

direction of motion.

Quiz: How much energy

does Earth spend in a

month in order to keep the

Moon on the orbit?

• In most cases, more than one force act on the moving objects

• In these cases the overall exchange of energy is measured using the net work: the

sum of all individual works which is the same with the work done by the net force

acting on the system.

• So, the calculation of the net work can be tackled in two ways:

1. first find the net force and then find the work it does, or

2. first find the work done by each force and add the works

Work – �et work done by several forces

Problem:Problem:

1. %et work on a flat surface: A box of mass m = 50 kg is pulled a distance d =10 m by a

force F =100 N making an angle θ =30o with the horizontal along a flat rough surface with

coefficient of kinetic friction µk= 0.1.

a) Calculate the work done by each force in the system

b) Determine the net work done on the box

F�

θ

d

m

Kinetic Energy – Definition

How can one use the idea of work to analyze motion?

• According to Newton’s 2nd Law, a non-zero net force applied to an object changes

its state of motion. Since the net force corresponds to a net work, we see that the

work measures the change in motion

• In order to see how we can associate work, energy and motion, let’s consider an

example: a box of mass m acted upon by a constant net force F such that its

acceleration a is constant and the velocity changes uniformly from v0 to v:

0r x x∆ = −� � �

F�

mHow is the net work done

on the box related to the

( )( )

0 2 21 10 02 22 2

0 0

2

W F r ma x xW mv mv K K K

v v a x x

= ⋅∆ = − ⇒ = − ≡ − = ∆

= + −

�

�

0 0,v x� �

0

,v x� �

212

K mv=• So, we see that the work done by the net force

results in a change of a quantity depending on

velocity and mass called kinetic energy K

on the box related to the

change in its motion?

• Even though we derived the expression for the how the work determines a change

in kinetic energy in a case with constant forces, it is a generic result:

net final initialW K K K= ∆ = −

Work-Energy Theorem: The net work done by a net force on a system will lead

in an equal change in the kinetic energy of the system:

• Notice that, if the net work is positive, the kinetic energy increases, whereas, if the

net work is negative, the kinetic energy decreases

Kinetic Energy – Relation to work

net work is negative, the kinetic energy decreases

• While the theorem is always valid (being a consequence of the conservation of

energy), the definition of kinetic energy works in the form on the previous slide only

for a an object in translational motion

Problem:

2. %et work down an incline: A box of mass m = 8.00 kg is initially at rest on top of an

incline of length d = 12 m and angle θ = 32°. The box is released down the incline which is

rough with coefficient of kinetic friction µk = 0.25.

a) Calculate the net work done on the box

b) Use the Work-Energy Theorem to find the speed of the box at the base of the incline.

Kinetic Energy – Elastic collisions

• An application of the concept of kinetic energy is in

elastic collisions where both momentum and kinetic

energy are conserved – since the net work done by

the forces exerted by the colliding objects upon each

other is overall zero

• Only for head-on elastic collisions, the conservation

of kinetic energy can be reduced to a simpler form:

2 2 2 21 1 1 1m v m vm v m v+ = +

− ′= − + ′

′ ′

v2

m2v1

m1

v′

before

collision

Wnet = 0

after

Problems:

3. Equations for head-on elastic collision: Demonstrate that the conservation of momentum

and kinetic energy in elastic collisions leads to the equations above.

4. Head-on elastic collision with stationary target: Two objects of masses m1 and m2 collide

head-on. Mass m1 has an initial speed v1, and mass m2 is initially at rest.

a) Calculate the speeds of the masses after collision in terms of the given quantities.

b) Comment on what is going to happen if m1 = m2, and if one the objects is much more

massive than the other

21 21v vv v− ′= − + ′1v′ 2v′

• So, we’ve seen that calculating the work done by a constant force is the dot product

between the force and the displacement. What if the force depends on the position?

• Let’s first approach the problem graphically: say that a force F acts upon a particle

moving in straight line along an x-axis. Graphically, we see the work is the area under

the force-vs-position graph

F

⇒

Work – Variable force. 1D motion

a) If F is constant: b) If F is not constant: F = F(x)

F

F4

F3

F2 ⇒

F

( )2

1

x

x

W F x dx= ∫F(x)

F(x)

1 1 2 2 ... i i

i

W F x F x F x= ∆ + ∆ + = ∆∑

x

( )2 1W F x x F x= − = ∆

1x 2x 1x∆ 2x∆ 3x∆ 4x∆

⇒

• Therefore, analytically, the work done by a force depending

on the position along a straight line is given by the integral of

the force along the straight path:( )

final

initial

x

x

x

W F x dx= ∫

x

F2

F1

⇒

x

1x 2xApproximation taking average

force in finite intervals:

If the intervals are infinitesimally

small, we cover the surface

F(x)

Problem:

5. Validity of the Work Energy Theorem: Demonstrate that the Work-Energy Theorem still

works in the case of forces varying along a straight line.

6. Work of a varying force: A box of mass m moves with speed v0 on a flat frictionless

surface, when a force starts acting on it. The force is parallel with the velocity, keeps the same

direction, and its magnitude depends on the position x with respect to the starting point as

given by F = Ax2 (where A is a constant).

a) Calculate the net work done on the particle as it moves a distance d

b) What is the speed of the particle after it travels the distance d?

•A canonical example of variable force appears in a deformed

spring, given by

eF kx= −�

�

appliedF

Work – Hooke’s Law. The work of spring forces

Hooke’s Law: If a spring is either compressed or

stretched through a certain elongation x with respect

to the equilibrium (un-stretched) position, the

deformation will be opposed by an elastic force Fe

proportional to the elongation:

x >0

x <0

x =0

k

k

0eF kx= − >

equilibrium

eF kx= −

appliedF

where k is called the force constant and characterizes

the stiffness of the spring

x >0

k

0eF kx= − <

• Therefore, the amount of work

done to stretch or compress a

spring a length x from equilibrium

x0 = 0 is given analytically and

then graphically by:

0 0

x x

xW F dx kxdx= =∫ ∫0 x

–kx

F

x

212

W kx⇒ =

• Consider a particle acted by a force F = (Fx, Fy),

moving along a path in the xy-plane made of many

elementary displacements dr

• Notice that the work between points AB is the integral

of the elementary works done on each elementary

displacement along the path, such that we have

F�

dr�

B B

A A

W dW F dr= = ⋅∫ ∫�

�

A

B

path

Work – Variable force. 2D motion

||F�

F⊥

�

• Therefore, in terms of vector components, we have

• In problems, the path of integration (trajectory) can be provided in various ways:

( ) ( )ˆ ˆ ˆ ˆx y x yW F F dx di y F dx F dj j yi= + ⋅ + = +∫ ∫ ∫

( ) ( ) ( )

final

initial

x

x y

x

dyy y x W F x F x dx

dx

= ⇒ = + ∫

( )( )

( ) ( )final

initial

t

x y

t

x x dx dyW F F d

dt d

tt t t

ty y t

= ⇒ = + = ∫

1. As a map in the xy-plane:

2. As a parametric curve in

terms of a certain parameter

t, such as a polar angle:

• Therefore, in terms of vector components, we have

ˆ ˆcos sinF A i A jθ θ= −�

y (m)

Problem:

7. Path given as a parametric curve: A particle P moves along a quarter of an arc with

radius R starting from the lowest end of the arc. A variable force is applied to the particle.

Using as a parameter the polar angle θ with respect to the positive x-axis shown on the figure,

the force is defined parametrically by

a) Write the path of the particle as a function of the parameter θ

b) Calculate the total work done by the force in terms of R and A

r�

P

θx (m)

y (m)

Power

•A transfer of energy specified using the concept of work doesn’t provide any

information about the rate of which the energy is transferred

• In order to specify how fast the energy can be released or consumed by a system

we use the concept of power

Ex: The difference between walking and running up the stairs is power – while the necessary

energy is the same (given by the change in gravitational potential energy) when running the

energy is spent faster: a larger power is necessary

Def: The rate of energy exchanged (or work W) by a system is called power:

• The average power can be written in terms of the force and velocity: if a certain

work is done onto an object by a force F, the power delivered instantaneously is

• This shows that, for a given power, a source of motion (such as an engine) will

instantaneously apply large forces at small speeds and vice-versa

dW F dxP F v

dt dt

⋅= = = ⋅

�

�

�

�

3

SI

1 J 4Watt (W) 10 hp

1 s 3P −= = ≈ ×

dWP

dt=

• The work done by some forces has special properties leading to a technique for

solving motions using energy based on the so called potential energy

• If a particle is under the influence of a certain force which can virtually do work –

such as weight, or elastic force –, but it is at rest due to a zero balance of forces, the

energy is stored in the system, it is only a potential energy

• So, the potential energy is associated with the configuration of the system, rather

than with motion.

• If the object is released, the potential energy can transform into kinetic energy.

• We’ll study two types of potential energies:

Potential Energy – Concept and types

• We’ll study two types of potential energies:

1. Gravitational potential energy,

Ug associated with the work done

by gravity

Def: The sum between kinetic energy and potential energy in a certain

configuration is called mechanical energy:

2. Elastic potential energy, Ue

associated with the work done by

elastic force in springs

E = K +Ug +Ue

( ) ( ) ( )0 0 0 gravity g g gW mg y y mgy mgy U U U= − − = − − = − − = −∆

• Consider a particle ascending vertically between two heights

y0 and y relative to an arbitrary ground, under the influence of

weight and some other forces doing work Wother

• Then, we see that the work done by gravity is:

Gravitational Potential Energy

Def: gravitational potential energy: y0

y

• Therefore, the work done by gravity does not depend of the

Ug = mgy

0other otherE EE W W∆ = ⇒ − =

( ) ( )210

212 02 othermv m Wm gyv mgy +−+ =

mg�

ground

• Therefore, the work done by gravity does not depend of the

path, but only on the final and initial positions

• Consequently, we can rewrite the Work-Energy Theorem:

• This result allows us to solve problems

using the concept of mechanical energy:

∆K = Wnet = Wgravity +Wother = −∆Ug +Wother

( )g other g otherK U W K U W∆ + ∆ = ⇒ ∆ + =

• In order to solve motion using mechanical energy, we have to write it out for

different configurations of the system and then use the conservation law

• For instance, if the only force acting on the moving object is its weight, the

mechanical energy is conserved, that is, the kinetic and potential parts must change

in order to keep the total constant:

Conservation of Mechanical Energy

212

const.mv mgy+ =

Comments:

kinetic potential

Comments:

• The gravitational potential energy is calculated with respect to an arbitrary ground

characterized by Ug = 0.

• We can choose the ground at any level, since the conservation of energy involves

only a difference of potential energies, which is the same in any system of

coordinates.

1y

ground

2y

y∆

1y′ground

2y′

yƥ In problems, it is a good idea to

choose the ground at the lowest

altitude in the respective problem

Exercise: Conservative Character of

Gravitational Work

We’ve based our argument about

potential energy by assuming that it does

not depend on the path. Consider a box

moving along the arbitrary path on the

figure, and show that the change in

gravitational potential energy depends

only on the final and initial altitudes.

( ) ( )gravity 2 1ˆ ˆ ˆW w s mgj xi yj mg y mg y y= ⋅∆ = − ⋅ ∆ + ∆ = − ∆ = − −

� �

• So, indeed, the work done by weight does not depend on the path: only on the initial and final • So, indeed, the work done by weight does not depend on the path: only on the initial and final

altitudes y1 and y2 (that is configurations of the system object-gravitational field)

• Later, we’ll see that this is a characteristics of the forces called conservative forces

Quiz:

Consider two frictionless ramps. If a block of mass m is released from rest at the left hand end

of each ramp, which block arrives at the right-hand end with the greater speed?

A B

Problem

8. Attwood machine revisited in an energy context: Two boxes with masses m1 and m2 are

connected by a massless cord passing over a light pulley, as in the figure. When the system is

released from rest (configuration A), the mass m1 moves upward and m2 downward a distance h

with respect to the initial position A.

a) Use the conservation of mechanical energy to derive an expression for the speed of the

system in configuration B.

b) Use the speed to find the acceleration of the system.

m1 m2

m1

m2

h

h

Elastic Potential Energy

• Like in the gravitational case, the work done by elastic forces depends only a final

and initial configuration of the system characterized by potential energies

x0

x

0eF�

eF�

• In order to find the potential energy of a spring for a

certain elongation, consider a spring compressed

from an initial compression x0 to a final one x

• The work will be given by the area of the graph

force F vs. elongation x which is simple to calculate

since F ~ x and the graph is a straight line:

k

k

k

( ) ( )2 2

x x

= − = − = − − = − − = −∆∫ ∫ ( ) ( )0 0

2 21 10 02 2

elastic e e e e

x x

W F dx kxdx kx kx U U U= − = − = − − = − − = −∆∫ ∫21

2eU kx=Def: elastic potential energy:

( )e other e otherK U W K U W∆ + ∆ = ⇒ ∆ + =

net elastic other e otherK W W W U W∆ = = + = −∆ +

• So, we can use the elastic potential energy as the

gravitational one, as part of the mechanical energy

• Consequently, the Work-Energy Theorem becomes:

• Consider an object of mass m connected to an ideal spring moving between

configurations (1) and (2) characterized by vertical positions y1,2 and spring

elongations x1,2

• If there are some “other” forces besides gravity and elastic forces, the mechanical

energy is not conserved:

• If there are no “other” forces, Wother = 0 and the mechanical energy is conserved:

Conservation of Mechanical Energy

( ) ( )2 2 2 21 1 1 12 2 2 1 1 12 2 2 2g e other otherK U U W mv kx mgy mv kx mgy W∆ + ∆ + ∆ = ⇒ + + − + + =

• In this expression, x is considered with respect to the equilibrium position of the

spring, and y with respect to an arbitrary ground

• This relationship allows us to analyze motions between different configurations: just

write the mechanical energy for each configuration that the system passes through and

then use the energy conservation to find the relations between speeds, positions, etc.

2 21 12 2

const.mv kx mgy+ + =2 2 2 21 1 1 12 2 2 1 1 12 2 2 2

mv kx mgy mv kx mgy+ + = + + ⇒

finalEinitialE K Ue Ug

Conservative and %onconservative Forces – Characteristics

• Why not consider the “other” forces the same way

we considered the gravitational and elastic forces by

defining potential energies? It is because we can only

associate potential energies with conservative forces

• Conservative forces do reversible work that depends

only on the final and initial configurations, whereas

nonconservative forces do path-dependent

dissipative work converted into internal energy Uint

Conservative %onconservative

• Gravitational

• Elastic

• Electric

• Friction

•Air resistance

• Tension in cords

• Push or pull by a

person

int int 0otherU W K U U∆ = − ⇒ ∆ + ∆ + ∆ =

Ex 2: %onconservative force: The work done

by the friction is larger when a crate is moved

between initial and final positions 1 → 2 along

a curved path instead of a direct straight line

f�

f�

f�

mg�

Ex 1: Conservative force: The work done

by gravity on a falling object is the same for

any path as long as the object travels

through the same difference of altitude

1

2 2

int intother

10. Elastic and gravitational potential energy: A 2.0 kg block is pushed against a spring

with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block

is released, it moves along a frictionless, horizontal surface and then up a frictionless incline

with slope 37.0°.

Problems:

9. Elastic potential energy: A 12-kg box is placed on a vertical spring of negligible mass and

force constant k = 1800 N/m that is compressed 15.0 cm. When the spring is released,

a) What is the speed of the box when its passes through the spring’s equilibrium position?

b) How high does the box rise from its initial position?

a) What is the speed of the block as it slides along the

horizontal surface after having left the spring?horizontal surface after having left the spring?

b) How far does the block travel up the incline before

starting to slide back down?

11. %onconservative work: A 0.50 kg block is pushed against a horizontal spring with

negligible mass and force constant k = 100 N/m, compressing it 0.20 m. When released, the

block moves on a horizontal table top for 1.00 m before coming to rest.

a) What is the coefficient of kinetic friction between the

block and the table?

b) What is the speed of the block when it leaves the

spring?

Problems

12. Inelastic collision and energy assessment: A 5.00 g bullet is fired horizontally into a

1.20 kg wooden block connected to a spring with force constant 100 N/m in equilibrium on a

horizontal surface with coefficient of kinetic friction µ = 0.25. The bullet remains embedded

in the block, and the composite system compresses the spring by x = 0.23 m before stopping.

What was the initial speed of the bullet?

Variation on the same theme: The problem can be reformulated

x

M km

Variation on the same theme: The problem can be reformulated

a) including only a rough surface

Calculate the maximum distance traveled.

θL

M

x

vm

b) including a pendulum

Calculate the maximum height y and the angle θ.

• So, conservative forces do work given by differences of potential energy

• Therefore, the potential energy given for a configuration x, U = U(x), can be used

to compute:

1. The work done while the configuration changed

2. The force acted instantaneously by the conservative field

Force and Potential energy

x x

dUdW F dx F

dx= ⇒ = −

dW dU= −

• We see that the force is given by the

1D:Ex: Elastic and gravitational forces are the

slopes of the respective potential energy curves

, ,dU dU dU

F Udx dy dz

= − ≡ −∇

� �

• We say that the force is given by the

“gradient” of the potential energy: this is

strictly a property of conservative forces

• We see that the force is given by the

negative of the slope of the potential energy

curve as a function of x

• The result can be extrapolated to 3D cases:

“del” or “nabla” operator

• Therefore, the graph U = U(x) can be used to analyze motion

• For instance, consider that the potential energy of a particle moving in a straight line

in a conservative field is given by an energy diagram:

Energy Diagrams

x

dUF

dx= −

• In any point the force is given by

• Maxima: unstable equilibrium

• Minima: stable equilibrium0

dU

dx=

• Minima: stable equilibrium

• If the total energy E = K + U is less than

two local successive maxima, the valley in

between is called a potential energy well

• The intersection points of E(x) with U(x)

are called turning points, since the particle

cannot exceed the maximum energy

• A potential energy well is flanked by

turning points so a particle would be

trapped in a well

dx

Problem:

13. Energy diagram: The figure shows the

potential energy diagram for a 20-g particle released

from rest at x = 1.0 m.

a) Draw the total energy line

b) Will the particle move to the right or to the left?

c) What is the particle’s maximum speed and where

does it reach it?

d) What are the most extreme positions that the

particle can reach – that is, where are the turning

points? points?

Power•A transfer of energy specified using the concept of work doesn’t provide any

information about the rate of which the energy is transferred

• In order to specify how fast the energy can be released or consumed by a system

we use the concept of power

Ex: The difference between walking and running up the stairs is power – while the necessary

energy is the same (given by the change in gravitational potential energy) when running the

energy is spent faster: a larger power is necessary

Def: The instantaneous time rate of energy exchanged (or work W done) by a

system is called power. If a finite time interval ∆t is considered, we calculate an

• The power can be written in terms of the force and velocity: for instance, in 1D, if a

certain elementary work dW is done onto an object by a force F parallel with motion,

the power delivered is

• This shows that, for a given power, the larger is the velocity of a certain motion , the

smaller is the force a source of motion is capable to impinge

x

dW dxP F Fv

dt dt= = =

system is called power. If a finite time interval ∆t is considered, we calculate an

average power

SIJ s Watt WP = = =

dWP

dt= netW

Pt

=∆