Chapter 7: Energy

Power: Sources of Energy

Tidal Power

SF Bay Tidal Power Project

Main Ideas(Encyclopedia of Physics)

Energy is an abstract quantity that an object is said to possess. It is not something you can directly observe. The usefulness of the concept comes from the Conservation of Energy. In predicting the behavior of objects, one uses the Conservation of Energy to keep track of the total energy and the transfer of energy between its various forms and between objects.

Work is the transfer of energy from one object to another by a force from one on the other that displaces the other.

Power is the rate at which energy is transferred or, the rate at which work is done. Power is the FLOW of energy.

Conservation of Energy

Energy can neither be created nor destroyed. It may change in form or be transferred from

one system to another.

The total amount of energy in the Universeis constant and can never change.

i fE E=

Except for VERY brief amounts of time according to the Heisenberg Uncertainty Principle.

Ways to Transfer Energy Into or Out of A System

• Work – transfers by applying a force and causing a displacement of the point of application of the force

• Mechanical Waves – allow a disturbance to propagate through a medium

• Heat – is driven by a temperature difference between two regions in space

More Ways to Transfer Energy Into or Out of A

System• Matter Transfer – matter

physically crosses the boundary of the system, carrying energy with it

• Electrical Transmission –transfer is by electric current

• Electromagnetic Radiation –energy is transferred by electromagnetic waves

Work Is An Energy Transfer

• If a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary.This will result in a change in the amount of energy stored in the system

• If the work is done on a system and it is positive, energy is transferred to the system

• If the work done on the system is negative, energy is transferred from the system

Units Energy & Work are scalars andhave units of the Joule:

2

2

mW Fd N m kg Js

= → ⋅ = ⋅ =

22 2

2

1 ( )2

m mKE mv kg kg Js s

= → = ⋅ =

Work A force applied across a distance.

cosθ= ΔW F r

Along the direction of motion ONLY! F must be parallel to the direction of motion!

•The sign of the work depends on the direction of F relative to Δr•Work is positive when projection of F onto Δr is in the same direction as the displacement•Work is negative when the projection is in the opposite direction

Work: W = F Δr cos θ– The displacement is that of the point of

application of the force– A force does no work on the object if

the force does not move through a displacement

– The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application

– The sign of the work depends on the direction of F relative to Δr

– Work is positive when projection of Fonto Δr is in the same direction as the displacement

– Work is negative when the projection is in the opposite direction 0fW <

0FW >

Kinetic EnergyThe energy an object has due to its motion.

212

KE mv=

IMPORTANT!v is the TOTALvelocity and is a scalar!!!

Table 7.1, p.194

Kinetic Energy

The net work done changes the Kinetic Energy.If the velocity is constant, then the net work is zero.

netW KE= Δ

Work-Energy Theorem

Notice: ( ) 2 21 12 2net f iW KE F r ma r mv mv= Δ → Δ = Δ = −∑

2 2 2f iv v a r= + Δ

rΔ

Quick Question (a)

A guy pushes on a 20kg mower with a force of 80 N at an angle of 25 degrees. How much work does he doing pushing the mower 50 meters?

cos 80 cos 25 50θ= Δ =W F r N x m3.63W kJ=

Quick Question (b)A guy pushes on a 20kg mower with a force of 80 N at an angle of 25 degrees. b) If starting from rest, what is the KE at the end of the 50 meters?

netW KE= Δ

f iKE KE= −

3.63fKE kJ=

) 3.63neta W kJ=

0

Quick Question (c)A guy pushes on a 20kg mower with a force of 80 N at an angle of 25 degrees. c) What is the speed of the mower at the end of the 50 m?

212

KE mv=

) 3.63neta W kJ=

) 3.63fb KE kJ=

2KEvm

=2 3,630

20J

kg⋅

=

19 /v m s=

Work-Energy Thm: HO 1

A 2.0-kg particle has an initial velocity of (5i – 4j) m/s. Some time later, its velocity is (7i + 3j) m/s. How much work was done by the resultant force during this time interval, assuming no energy is lost in the process?a. 17 Jb. 49 Jc. 19 Jd. 53 Je. 27 J

netW KE= Δ

Energy StatesLeft to their own devices, systems always seek out the

lowest energy state available to them.Systems want to be at rest or in a constant state of

motion.

You have to do work on the Rock to roll it back up the hill. This will give the Rock Energy – the potential of rolling back down – Potential Energy.

Potential EnergyThe energy an object has due to its position in a force field.

For example: gravity or electricityThe Potential Energy is relative to a ‘ground’ that is defined.

(Potential Energy: U or PE)

Gravitational Potential EnergyPE mgh=Force Distance

It takes work to move the object and that gives it energy!

IMPORTANT!Either path gives the

same potential energy!WHY?

Same change in height!

PE = mghThe “ground”: h = 0

Work Up an InclineThe block of ice weighs 500 Newtons. How much work does it take

to push it up the incline compared to lifting it straight up?Ignore friction.

Work Up an InclineWork = Force x Distance

mg = 500N

Straight up: 500 3 1500W Fd N m J= = ⋅ =

250 6 1500W Fd N m J= = ⋅ =

?F = sinF mgPush up:

An incline is a simple machine!

θ=

What is the PE at the top? 1500J

3500 2506

N N= ⋅ =

Simple Machines

Same Work, Different Force, Different Distance

Force Multipliers

Use Scalar Product of Two Vectors!

• The scalar product of two vectors is written as A . B– It is also called the

dot product

• A . B = A B cos θ−θ is the angle

between A and Bcosθ= ΔW F r

= ⋅ Δv vW F r

x x y y z zA B A B A B A B⋅ = + +

1cos ( )A BAB

θ − ⋅=

v v Becomes…

ProblemA force acts on a particle that undergoes a displacement

Find (a) the work done by the force on the particle and (b) the angle between F and r.

F = 6ˆ i – 2ˆ j

Unit Vector Representation.How do you find the Work?

( )N

Δr = 3ˆ i + ˆ j ( )m

Problem F = 6ˆ i – 2ˆ j ( )N Δr = 3ˆ i + ˆ j ( )m

= ⋅ Δ = +F r x yW F x F y

1cosθ − ⋅ Δ⎛ ⎞= ⎜ ⎟Δ⎝ ⎠

F rF r

( )( ) ( )( )6.00 3.00 N m 2.00 1.00 N m 16.0 J= ⋅ + − ⋅ =

( ) ( )( ) ( ) ( )( )1

2 2 2 2

16cos 36.96.00 2.00 3.00 1.00

−= = °+ − +

Find (a) the work done by the force on the particle and (b) the angle between F and r.

a)

b)

You Try HO5If the resultant force acting on a 2.0-kg object is equal to (3i + 4j) N, what is the change in kinetic energy as the object moves from(7i – 8j) m to (11i – 5j) m?a. +36 Jb. +28 Jc. +32 Jd. +24 Je. +60 J

Stopping Distance

NetW KE= Δ 212

Δ =F r mv21 /

2Δ =r mv F

Traveling at 70 miles per hour, what is your breaking distance?

2

2Δ =

vra

If v doubles, d quadruples!!!

221 /

2 2= =

vmv maa

Problem

Suppose you push on a 30.0kg package initially at rest with a force of 120.0 N through a distance of 0.800m

against an opposing friction of 5.00N. What is the kinetic energy of the box at the end of the 0.80 m?

0 0v =

ProblemnetW KE= Δ

( )

Use Work-Energy Theorem:

f iF f d KE KE− ⋅ = −

0 0v =

( )fKE F f d= − ⋅ = ( )120 5 0.8 92N N m J− ⋅ =

net netW F d KE= = Δ

Problem

0 0v =

92KE J=What is the final velocity of the box?

212

KE mv= 2KEvm

=2 9230

Jkg

⋅=

2.48 /v m s=

HO 6

A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P = 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B? The angle of the incline is 30 degrees.a. 27 Jb. 20 Jc. 24 Jd. 17 Je. 37 J

3030˚

AP

B

YOU TRY HO2A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 Jand 5.6 J, respectively, how much work is done on the block by the force P between A and B?

P

A B

a. 2.7 Jb. 1.0 Jc. 2.2 Jd. 1.6 Je. 3.2 J

Work Done by a Varying Force

The work done is equal to the area under the curve

f

i

x

xxW F dx= ∫

cosθ=x

whereF F

H04A force acting on an object moving along the x axis is

given by Fx = (14x – 3.0x2) Nwhere x is in m. How much work is done by this force as the object moves from x = –1 m to x = +2 m?a. +12 Jb. +28 Jc. +40 Jd. +42 Je. –28 J

You Try HO3

An object moving along the x axis is acted upon by a force Fxthat varies with position as shown. How much work is done by this force as the object moves from x = 2 m to x = 8 m?a. –10 Jb. +10 Jc. +30 Jd. –30 Je. +40 J

10

0

–10

20

Fx (N)

x (m)642 8 10 12642 10 128

Work Done by a Varying Force: Gravity

22

12 2

1.3 10xFr

= −

The interplanetary probe is attracted to the sun by a force given by:

The negative sign indicates that the force is attractive. This is because of the way that the polar unit vectors are defined. With the origin located at the sun and the radial vector pointing towards the probe, the force of gravity acting on the probe is in the negative direction.

Probe

Sun

Work Done by a Varying Force: Gravity

11

11

222.3 10

21.5 10

1.3 10x

x

xW dxx

⎛ ⎞= −⎜ ⎟

⎝ ⎠∫

The probe is moving away from the sun so the work done ON the probe BY the sun is slowing it down. Thus, the work should be negative.

11

11

2.3 1022

1.5 10

11.3 10 ( )x

xx x−= − −

103 10x J= −

Attractive force versus distance for interplanetary probe.The area under the curve is negative since curve is below x-axis.

22

12 2

1.3 10xFr

= −

Hooke’s Law Fs = - kx

• The restoring force exerted by the spring is Fs = - kx

• x is the position of the block with respect to the equilibrium position (x = 0)

• k is called the spring constant or force constant and measures the stiffness of the spring

Work Done by a Varying Force

Hooke’s Law

An elastic system displaced from equilibrium oscillates in a simple way about its equilibrium position with

Simple Harmonic Motion.

Hooke’s Law describes the elastic response to an applied force.

Elasticity is the property of an object or material which causesit to be restored to its original shape after distortion.

Ut tensio, sic vis - as the extension, so is the force

Elastic SystemsSmall Vibrations

F kx= −

Robert Hooke (1635-1703)

•Leading figure in Scientific Revolution•Contemporary and arch enemy of Newton•Hooke’s Law of elasticity•Worked in Physics, Biology, Meteorology, Paleontology•Devised compound microscope•Coined the term “cell”

• When x is positive (spring is stretched), F is negative

• When x is 0 (at the equilibrium position), F is 0

• When x is negative (spring is compressed), F is positive

Hooke’s Law Fs = - kxThe Restoring Force

Hooke’s LawIt takes twice as much force to stretch a spring twice as far.

The linear dependence of displacement upon stretching force:

appliedF kx=

Hooke’s LawStress is directly proportional to strain.

( ) ( )appliedF stress kx strain=

Hooke’s Law

•The applied force displaces the system a distance x.

•The reaction force of the spring is called the “Restoring Force” and it is in the opposite direction to the displacement.

RestoringF kx= −

Spring Constant k: Stiffness

•The larger k, the stiffer the spring•Shorter springs are stiffer springs•k strength is inversely proportional to the number of coils

Spring Question

Each spring is identical with the same spring constant, k.Each box is displaced by the same amount and released.Which box, if either, experiences the greater net force?

Work Done by a Varying Force

The work done is equal to the area under the curve

f

i

x

xxW F dx= ∫

cosθ=x

whereF F

Conservative Forces• The work done by a

conservative force on a particle moving between any two points is independent of the path taken by the particle.

Examples: Gravity, Spring force & Electricity

• The work done by a conservative force on a particle moving through any closed path is zero– A closed path is one in which

the beginning and ending points are the same

Conservative Forces and Potential Energy

• Define a potential energy function, U, such that the work done by a conservative force equals the decrease in the potential energy of the system

• The work done by such a force, F, is

• For an infinitessimal displacement:

f

i

x

C xxW F dx U= = −Δ∫

xdUFdx

= −

Conservative Forces and Potential Energy – Check

• Look at the case of a deformed spring

– This is Hooke’s Law

– Gravitational Potential & Force:

212

ss

dU dF kx kxdx dx

⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

( )= − = − = −gg

dU dF mgy mgdx dx

xdUFdx

= −

Energy Diagrams and Stable Equilibrium: Mass on a Spring

• The x = 0 position is one of stable equilibrium

• Configurations of stable equilibrium correspond to those for which U(x) is a minimum.

• x=xmax and x=-xmax are called the turning points

xdUFdx

= −

Energy Diagrams and Unstable Equilibrium

• Fx = 0 at x = 0, so the particle is in equilibrium

• For any other value of x, the particle moves away from the equilibrium position

• This is an example of unstable equilibrium

• Configurations of unstable equilibrium correspond to those for which U(x) is a maximum. Ex: A pencil standing on its end.

xdUFdx

= −

P7.47For the potential energy curve shown, (a) determine whether the force Fx is positive, negative, or zero at the five points indicated. (b) Indicate points of stable, unstable, and neutral equilibrium. (c) Sketch the curve for Fxversus x from x = 0 to x = 9.5 m.

A

B

C

D

E x (m)

Fx

a) Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.

xdUFdx

= −

b) A and E are unstable, and C is stable.

Work Done by a Spring

• Identify the block as the system

• The work is the area under the

• Calculate the work as the block moves from xi = - xmax to xf = 0

• The total work done as the block moves from –xmax to xmax is zero.

( )max

0 2max

12

f

i

x

s xx xW F dx kx dx kx

−= = − =∫ ∫

Conservative Forces and Potential Energy – Check

• Look at the case of a deformed spring

– This is Hooke’s Law

– Gravitational Potential & Force:

212

ss

dU dF kx kxdx dx

⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

( )= − = − = −gg

dU dF mgy mgdx dx

xdUFdx

= −

Spring with an Applied Force

• Suppose an external agent, Fapp, stretches the spring

• The applied force is equal and opposite to the spring force

• Fapp = -Fs = -(-kx) = kx

• Work done by Fapp is equal to ½ kx2

max

Energy in a SpringWhat speed will a 25g ball be shot out of a toy gun if the spring (spring constant = 50.0N/m) is compressed 0.15m? Ignore friction and the mass of the spring.

Use Energy!

= Δspring ballW KE

2 21 12 2

kx mv=

kv xm

=

50.0 / (.15 ) 6.7 /.025

N mv m m skg

= =

You Try HO 9

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? k = 2.0 kN/m The mass of t he block is 2.0 kg.a. 3.7 m/sb. 4.4 m/sc. 4.9 m/sd. 5.4 m/se. 14 m/s

2.0 kg2.0 kg2.0 kg

k

v

HO 10A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1.4 kN/m). The block is pulled 8.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?a. 4.5 Jb. 2.1 Jc. 6.9 Jd. 6.6 Je. 4.9 J