Work, Powerand Conservation of Energy

Forms of Energy

• Mechanical• Chemical/Thermal/Electrochemical• Electromagnetic• Nuclear• Energy can be transformed from one form to

another– Essential to the study of physics, chemistry,

biology, geology, astronomy• Can be used in place of Newton’s laws to solve

certain problems more simply

Transferring Energy

• By Work– By applying a force– Produces a

displacement of the system

Transferring Energy

• Heat– The process of

transferring heat by collisions between molecules

Transferring Energy

• Mechanical Waves– a disturbance

propagates through a medium

– Examples include sound, water, seismic

Transferring Energy

• Electrical transmission– transfer by means

of electrical current

Transferring Energy

• Electromagnetic radiation– any form of

electromagnetic waves• Light,

microwaves, radio waves

Work

• Provides a link between force and energy• The work, W, done by a constant force on an

object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement

θ

Facosθ

F a Fasinθ

displacement

Work

• • F cos θ is the component of

the force in the direction of the

displacement• Δ x is the displacement• Note that if the distance is zero (if the box doesn’t

move) then the work is zero.• This gives no information about

– the time it took for the displacement to occur– the velocity or acceleration of the object

Units of Work

• Work = Force x distance Joule = Newton x meter

• The Joule is the unit for work.• The Joule is the unit for energy.

More About Work• Scalar quantity (so negative signs matter!)• The work done by a force is zero when the

force is perpendicular to the displacement– F cos 90° = 0 F Δx

• If there are multiple forces acting on an object, the total work done is the algebraic sum of the amount of work done by each force (in the direction of motion).

When Work is Zero

• Displacement is horizontal

• Force is vertical• cos 90° = 0

Work when force is perpendicular to motion

• For uniform circular motion, the force (Fc ) is always perpendicular to the motion (velocity). So this kind of force only causes a change in direction, it doesn’t change the speed.

• If speed doesn’t change then ΔKE = 0.• If ΔKE = 0 then W = 0• So to repeat, a force applied perpendicular to

the motion results in zero work being done.

v Fc

More About Work, cont.

• Work can be positive or negative– Positive if the force and the displacement are in

the same direction– Negative if the force and the displacement are in

the opposite direction

Work Can Be Positive or Negative• Work is positive when lifting the box

– It takes effort to lift the box– The box gains potential energy when

you lift it. Energy gain takes work.• Work would be negative if lowering the box

– Your force is 1800 from the motion and cos 180 = -1

– The box would fall even if you didn’t lower it.

– The box loses potential energy when you lower it so energy loss = negative work

Determining work done by a forceEXAMPLE: A pulley system is used to raise a 100-N crate 4 m as shown. Find the work done by the tension force T if the lift occurs at constant speed.SOLUTION:• From the FBD since a = 0, T = 100 N.• From the statement of the problem, s = 4 m.• Since the displacement and the tension are

parallel, = 0°.• Thus W = Ts cos = (100)(4) cos 0° = 400 J.

Work, energy, and power - pulleys

T

TT

s

s

FYI• Pulleys are used to redirect tension forces.

T

FBD CrateT

100

a = 0

Determining work done by a forceEXAMPLE: A pulley system is used to raise a 100-N crate 4 m as shown. Find the work done by the tension force T if the lift occurs at constant speed.SOLUTION:• From the FBD 2T = 100 so that T = 50 n.• From the statement of the problem, s = 4 m.• Since the displacement and the tension are

parallel, = 0°.• So W = T(2s) cos = (50)(24) cos 0° = 400 J.

Work, energy, and power - pulleys

FYI• Pulleys are also used gain mechanical advantage.

T

T

T

2s

s

T T

FBD CrateT

100

a = 0

T

M.A. = Fout / Fin = 100 / 50 = 2.

+ or – sign for work• Let’s say that another way• If the external world does work on the system,

then work is positive and the energy in the system increases.

• If the system expends work on the external world, then work is negative and the energy in the system decreases.

• This will most likely be a source of confusion for you. Plan to THINK about it carefully.

Work done by friction

• Work done by friction is acting at 180 degrees relative to the direction of motion.

• Since cos (180) = -1, the work done by friction is always negative.

• Negative work means it reduces the energy of the system (friction slows the object down).

Example 1

• An 105 g hockey puck slides across the ice. A player’s hockey stick exerts a constant 4.5 N force over a distance of 0.15 m. How much work does the player do on the puck? What is the change in the puck’s energy?

F

d

Example 2

• A sailor pulls a boat a distance of 30m along a dock using a rope that makes a 25 degree angle with the horizontal. Fapplied = 255 N

• How much work is done?

Graphical interpretation of work• A graph of force vs.

displacement allows you to find the work done.

• Work = F d = area under curve

• Top example:– W = (20N)(1.5m) = 30 J

• Bottom example:– W = ½ (20 N)(1.5m) = 15 J

Kinetic Energy• Kinetic energy is associated with the motion of an object.• • It is a scalar quantity with the same units as work, Joules.• Work is related to kinetic energy by the Work-Energy Theorem • When work is done by a net force on an object and the

only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy.– Speed will increase if work is positive– Speed will decrease if work is negative

Derivation of the Work-Energy Theorem

• Kinematics Eq 4

• now mult. both sides by m/2

• Work = KEf - KEi

Work and Kinetic Energy• An object’s kinetic energy

can also be thought of as the amount of work the moving object could do in coming to rest

• The moving hammer has kinetic energy and can do work on the nail by giving up some of its kinetic energy.

Power • Power is equal to the work done

divided by the time taken to do that work.

• P = W/t• Power is measured in watts. • 1 Watt = 1 Joule/sec• Let’s say you have two people with the same mass

going up the same flight of stairs. One of them runs up the stairs and the other one walks. Both do the same amount of work, but the one

running has more power because her time is less

Power Example• An electric motor lifts an elevator

9.00 m in 15.0 sec by exerting an upward force of 1.2 x 104 N. What power does the motor produce in kW?

• P = W/t• P = Fd/t• P = (1.2 x 104 N)(9.00m) / 15.0 sec• P = 7200 W or 7.20 kW

Alternative (and always forgotten) expression for power

• P = W/t Power = Work/Time• P = Fd/t• P = (F)(d/t)• P = F v Power = Force x velocity

• So power can also be:• Power = Force x velocity

Power, cont.• US Customary units are generally hp

– need a conversion factor

– Can define units of work or energy in terms of units of power:• Energy = power x time• Energy = kilowatts x hours = kWh• kilowatt hours (kWh) are often used in electric

bills.• Note that kWh is a unit of energy NOT power!

Forms of Energy

Gravitational Potential Energy• Gravitational Potential Energy is the energy

associated with the relative position of an object in space near the Earth’s surface.

• All PE is measured relative to a reference level.• The reference level is often taken to be ground level.• Objects interact with the earth through gravitational

force• PE is actually the potential energy of the earth-object

system

• PE = mgh• Units are Joules, just like for KE and Work

Reference Levels for Gravitational Potential Energy

• A location where the gravitational potential energy is zero must be chosen for each problem– The choice is arbitrary since the change in the

potential energy is the important quantity– Choose a convenient location for the zero reference

height• Often it is the Earth’s surface (ground level)• May be some other point suggested by the

problem (table top, etc.)– What is the nature of potential energy? It is basically

stored energy that when released, causes motion.

Example – 1 kg ball Reference Level = the ground

Etotal = PE + KE

Etotal = mgh + 0 = (1)(9.8)(1) = 9.8 J

Etotal = PE + KE vf2 = vi

2 + 2aΔy

Etotal = mgh + ½ mv2 vf2 = 0 + 2(-9.8)(-0.5)

Etotal = (1)(9.8)(0.5) + ½ (1)(9.8) << vf2 = 9.8 m/s

= 9.8 J

Etotal = PE + KE vf2 = vi

2 + 2aΔy

Etotal = 0 + ½ mv2 vf2 = 0 + 2(-9.8)(-1)

Etotal = 0 + ½ (1) (19.6) = 9.8 J <<vf2 = 19.6 m/s

1m

v=0

0.5m

Relationship between KE and PE

Note how the sum of KE + PE doesn’t change. It stays the same due to conservation of energy.

Work and Gravitational Potential Energy

• PE = mgh• • That’s the way Serway defines it, but be

careful because that’s nonstandard relative to the way

we normally define Δ• Note that what he’s really saying is • Units of Potential Energy are the same as those of

Work and Kinetic Energy, Joules

Example• You lift a 7.3 kg bowling ball from

the storage rack and hold it up to your shoulder. The storage rack is 0.610m above the floor and your shoulder is 1.12 m above the floor.

When the bowling ball is at your shoulder, what is the ball’s gravitational PE relative to the floor?

• PE = mgh = (7.3kg)(9.8 m/s2)(1.12m) = 80.1 J

When the ball is at your shoulder, what is its PE relative to the rack? PE = mgh = (7.3kg)(9.8 m/s2)(1.12 – 0.610m) = 36.5 J

How much work was done by gravity as you lifted the ball from the rack to your shoulder?

W = Fd= (-mg)(d) = (-7.3kg)(9.8 m/s2 )(1.12-0.61) = -36.5 J

Example 2• You slowly (a=0) lower a 20 kg bag of sand 1.2 m

from the trunk of the car to the driveway. How much work do you do? Fa = mg Fa

• W = F d = mgh a=0• = 20(9.8)(0-1.2) = -235 J mg• The work is negative because the height decreased

from 1.2 to 0, meaning it lost potential energy.• If the system lost PE, then the work is negative.• Evidence? If you simply dropped the bag from the

height of 1.2 m it would have fallen to the ground anyway.

Example 3

• If a 1.8 kg brick falls to the ground from a chimney that is 6.7 m high, what is the change in PE?

• ΔPE = mgΔh = (1.8)(9.8)(0-6.7) = -120 J• The brick loses PE as it falls, so the number is

negative. • (Remember, PE is zero when it hits the

ground)

Potential Energy Stored in a Spring• Involves the spring constant k• Hooke’s Law gives the force

F = - k x• F is the restoring force• F is in the opposite direction of x• k is a measure of how stiff the spring is (high

numbers mean it’s hard to compress/stretch it, low numbers mean it’s easy to compress/stretch it.)

PEspring = ½ kx2 write on notecard!

Elastic Potential Energy• A bow and arrow set up stores elastic potential

energy. When you pull on a bow string you do work on that system (meaning you put energy into the system). That becomes stored energy in the form of elastic potential energy. When the arrow is released, that stored elastic potential energy is converted into the kinetic energy of the arrow.

• Elastic potential energy is stored in things like rubber bands, slingshots, springs, and trampolines.

• PEelastic = PEspring = ½ kx2

Law of Conservation of Energy• Conservative forces: a force is conservative if

the work it does on an object moving between two points is independent of the path the object takes between two points.– Example: gravity

• Nonconservative forces: A force is nonconservative if the work it does on an object depends on the path taken by the object between its final and starting points. – Example: friction

Conservative forces

• For a conservative force like gravity, it doesn’t matter what path you take to get from point A to point B, the energy involved is the same.

• Does that seem surprising to you?

a

B

Nonconservative force

• Nonconservative force – Friction means that

the book will lose more energy if it goes on the gold colored path then if it goes on the blue path.

Conservation of Mechanical Energy• Conservation in general

– To say a physical quantity is conserved is to say that the numerical value of the (total) quantity remains constant

KEi + PEi = KEf + PEf

• In Conservation of Energy, the total mechanical energy remains constant– In any isolated system of objects that interact only

through conservative forces, the total mechanical energy of the system remains constant.

KE and PE Animation

ch11_scn35_smallanim.swf

KE and PE, negative displacement

ch11_scn39_smallanim.swf

Conservation of Energy

• The law of conservation of energy states that in a closed, isolated system, energy can neither be created or destroyed, it only changes form.

• Mechanical energy = KE + PE (if no other forms of energy are present)

• Therefore an increase in KE = decrease in PE of the same amount.

Problem Solving with Conservation of Energy

• Define the system, verify it is a closed system.• Select the location of zero gravitational potential energy

(the reference level)– Do not change this location while solving the problem

• Identify the forms of energy (PE, KE, PEspring )• Determine whether or not nonconservative forces are

present– If there are no external forces, then

Etotal before = Etotal after

– If there are external forces doing work then: Etotal before + Wadded to system = Etotal after

or Etotal before - Wlost by system = Etotal after

Pendulums

• A the top of the swing (A or C), the energy is all PE, because the velocity >> 0 at the top of the swing.

• At the bottom of the swing, the energy is pure KE, because the PE is at its lowest point (the reference level where we let PE =0)

• The orange line shows the PE. Where is the PE maximum?

• The light blue line shows the KE. Where is the KE maximum?

• Where does the pendulum have it’s maximum speed? Minimum speed?

Example 1 - review• Let’s say you had a 10 N ball, 2 m above

the ground. Then the potential energy is:

• PE = mgh = (10 N) (2 m) = 20 Joules• If we drop the ball and it hits the ground,

what will be its Kinetic Energy in the instant just before it hits the ground?

• 20 Joules, because all of the PE has been converted to KE

• What would the KE be when the ball is 1m above the ground?

• 10 Joules, because half of the initial PE of 20 J has been lost (it’s height has been cut in half).

0.0 J

20.0 J

Example 2

• Can the ball get from point A to B?

A

B

Example 3

Example 3 (prob 2 p. 296 Zitzewitz)

• During a hurricane, a large tree limb, with a mass of 22 kg and a height of 13.3 m above the ground, falls on a roof that is 6.0 m above the ground. a) Find the KE of the limb when it reaches the roof. You could

find it from kinematic equation 4, but it is easier to find it from PE

rearranging, and using KEi = 0

b) Find the speed of the limb when it reaches the roof.

so and

Collisions : Case 1• If a system is isolated, then momentum and energy are conserved

• Check: pi = (1)(1) + (1)(0) =1 pf = (1)(-0.2) + (1)(1.2) = 1 • So momentum is conserved.• KEi = ½ (1)(1)2 + ½ (1)(0)2 = 0.5 J

• KEf = ½(1)(-0.2)2 + ½ (1)(1.2)2 = 0.74J

• Kinetic energy increased. When the two carts collided, maybe a spring was released, adding KE. That would be called a superelastic or explosive collision.

Collisions : Case 2

pi = (1)(1) + (1)(0) = 1 N s pf = (1)(0) + (1)(1) = 1 N sSo momentum is conserved KEi = ½ (1)(1)2 + ½(1)(0)2 = 0.5 JKEf = ½ (1) (0)2 + ½ (1)(1)2 = 0.5 JSo in this case, energy is conserved and KEi = KEf

If energy is conserved, this is called an elastic collision

Collisions: Case 3

Let’s check this case:pi = (1)(1) + (1)(0) = 1 N s pf = (1+1)(0.5) = 1 N sSo momentum is conserved in this perfectly inelastic collision

KEi = 0.5 J (same as Case 1 and 2) KEf = ½ (1+1) (.5)2 = 0.25 J

So this collision loses energy (goes from 0.5 J to 0.25 J).So this is an inelastic collision, where energy is lost.

Three different collision cases summary

• Superelastic or explosive collision

• Elastic collision

• Inelastic collision

Example 4 (p. 299 Zitzewitz)

mA = 575 kg mB = 1575vA = 15 m/s vB = 5 m/sfind vf and ΔKE.

so ΔKE = KEf – KEi = -2.12 x 104 Joule. Energy was lost.

Example 5 – Cons. of Energy

• Starting from vi = 0, a child zooms down a frictionless slide (to the ground) from an initial height of 3.00 m. What is her vf ? Assume a mass of 25.0 kg.

Note that all m’s in this equation cancel out 0 + (9.8)(3.00) = ½ vf

2 + 0

vf = 7.67 m/s

Example 6 – Cons. of Energy

• The spring on a jack-in-the-box is compressed 8 cm then released. What is the speed of the toy head when it returns to its natural state? (xf = 0). Given mhead = 0.05 kg, k=80 N/m and the motion is only vertical.

KEi + PEgi + PEsi = KEf + PEgf + PEsf

½ mvi2 + mg(hi – hf) + ½ kxi 2 = ½ mvf

2 + ½ kxf2

0 + (0.05)(9.8)(-0.08) + ½ (80)(-.08)2 = ½ (.05)vf2 + 0

-0.0392 + 0.256 = 0.025 vf2

Vf = 2.9 m/s

Example 7 – Cons. of Energy• Desperado, a roller coaster in Nevada, has a vertical drop of

68.6m. The roller coaster is designed so that the speed of the cars at the end of the drop is 35.6 m/s. Assume the carts are at rest at the top of the drop. What % of the initial mechanical energy is dissipated by friction?

KEi + PEi = KEf + PEf

PEi - PEf = KEf

mgΔh = ½ mvf2 the m’s cancel out

(9.8)(68.6) = ½ vf2

vf = 36.7 m/s if it had no friction losses

%Mech. Energy lost = ½ mvf 2no friction - ½ m vf 2

with friction

½ m vf 2no friction

% Mech. Energy lost = (36.72 – 35.62) / 36.72 = 5.9 %

Example 8 – Power/Work/PE

• Suppose someone threw a frisbee that stayed in the air for 16.7 seconds. Suppose this person ran up a staircase to catch the frisbee during this time, reaching a height of 18.4 m. If his mass was 72 kg, how much power was needed for his ascent?

• W = gain in PE = mgh• P = W/t• P = mgh/t• P = (72)(9.8)(18.4) / (16.7) = 778W.

FrFsW

rs

Fr

W

Rotational Work and Energy

Rotational Work and Energy

DEFINITION OF ROTATIONAL WORK

The rotational work done by a constant torque in turning an object through an angle is

RW

Requirement: The angle mustbe expressed in radians.

SI Unit of Rotational Work: joule (J)

Rotational Work and Energy

22122

2122

21 ImrmrKE

22212

21 mrmvKE T

rvT

Rotational Work and Energy

221 IKER

DEFINITION OF ROTATIONAL KINETIC ENERGY

The rotational kinetic energy of a rigid rotating object is

Requirement: The angular speed mustbe expressed in rad/s.

SI Unit of Rotational Kinetic Energy: joule (J)

Rotational Work and Energy

ExampleA thin-walled hollow cylinder (mass = mh, radius = rh) anda solid cylinder (mass = ms, radius = rs) start from rest atthe top of an incline.

Determine which cylinder has the greatest translationalspeed vf upon reaching the bottom.

Rotational Work and Energy

mghImvE 2212

21

iiifff mghImvmghImv 2212

212

212

21

iff mghImv 2212

21

USE ENERGY CONSERVATION

rv ff

Since ωi vi and hf are zero,

iff mghrvImv 22212

21

2

2

rIm

mghv of

RESULT: The cylinder with the smaller moment of inertia will have a greater final translational speed.

Rotational Work and Energy

• To summarize:• The rotational work WR done by a constant torque τ in turning

an object through an angle θ is • and work is expressed in J.• Rotational kinetic energy is derived from KE = Σ½mv2 and

since v = rω, then KE = Σ½ mr2ω 2 = ½(Σmr2)ω 2

• So rotational kinetic energy is • also expressed in units of J.• So keep in mind that for a case that has rotational and

translational energy:

Circus Energy – Russian Acrobats• http://www.pbs.org/opb/circus/classroom/circus-physics/conservation-energy/• 2’ video on kinetic and rotational energy