Work and Energy Dr. Robert MacKay Clark College. Introduction What is Energy? What are some of the...

Work and EnergyWork and Energy

Dr. Robert MacKay

Clark College

Introduction Introduction

What is Energy? What are some of the different forms of

energy? Energy = $$$

Overview Overview Work (W) Kinetic Energy (KE)

Potential Energy (PE) All Are measured in Units of Joules (J) 1.0 Joule = 1.0 N m

W KE

PE

Overview Overview

Work Kinetic Energy Potential Energy

W KE

PE Heat LossHeat Loss

Heat Loss

Crib SheetCrib Sheet

W FD// (JOULES)

KE 12mv2

GPE mgh

SPE 1

2kx 2 F kx

P W

t(J / sWatt )

Wnet KE

E KE PE

E f E0 WNC


Work = Force x distance W = F d Actually Work = Force x Distance parallel to force

d=4.0 m

F= 6.0 N

W= F d = 6.0 N (4.0m) = 24.0 J


Work = Force x Distance parallel to force

d= 8.0 m

F= 10.0 N

W = ?



d= 8.0 m

F= 10.0 N

W = 80 J



d= 8.0 m

F= - 6.0 N

W= F d = -6.0 N (8.0m) =-48 J



d= 6.0 m

F= - 5.0 N

W= F d = ? J



d= 6.0 m

F= - 5.0 N

W= F d = -30 J



d= 6.0 m

F= ? N

W= 60 J



d= 6.0 m

F= 10 N

W= 60 J



d= ? m

F= - 50.0 N

W= 200 J



d= -4.0 m

F= - 50.0 N

W= 200 J



d= 8.0 m

F= + 6.0 N

W= 0(since F and d are perpendicular

PowerPower

Work = Power x time 1 Watt= 1 J/s 1 J = 1 Watt x 1 sec 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J Energy = $$$$$$ 1 kW-hr = $0.08 = 8 cents

Power Worktime

J / s

PowerPower Work = Power x time W=P t [ J=(J/s) s= Watt * sec ]

work = ? when 2000 watts of power are delivered

for 4.0 sec.

PowerPower Work = Power x time W=P t [ J=(J/s) s= Watt * sec ]

work = 8000J when 2000 watts of power are delivered

for 4.0 sec.

PowerPower Energy = Power x time E =P t [ kW-hr=(kW) hr] or [ J=(J/s) s= Watt * sec ]

PowerPower Energy = Power x time

How much energy is consumed by a 100 Watt lightbulb when left on for 24 hours?

What units should we use? J,W, & sor kW-hr, kW, hr


How much energy is consumed by a 100 Watt lightbulb when left on for 24 hours?


Energy=0.1 kWatt (24 hrs)=2.4 kWatt-hr


What is the power output of a duck who does 3000 J of work in 0.5 sec?



What is the power output of a duck who does 3000 J of work in 0.5 sec?power=energy/time =3000 J/0.5 sec =6000 Watts


PowerPower Energy = Power x time E =P t [ kW-hr=(kW) hr]

Energy = ? when 2000 watts (2 kW) of power are

delivered for 6.0 hr.

Cost at 8 cent per kW-hr?

PowerPower Energy = Power x time E =P t [ kW-hr=(kW) hr]

Energy = 2kW(6 hr)=12 kW-hr when 2000 watts (2 kW) of power are delivered for

6.0 hr.

Cost at 8 cent per kW-hr? 12 kW-hr*$0.08/kW-hr=$0.96

MachinesMachines

Levers D =8 md = 1 m

f=10 NF=?

Work in = Work out

f D = F d

The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

MachinesMachines

Levers D =8 md = 1 m

f=10 NF=?

Work in = Work out 10N 8m = F 1m

F = 80 N


MachinesMachines

Pulleys

Dd

f

F

Work in = Work out

f D = F d


MachinesMachines

Pulleys

Dd

f

F

Work in = Work out

f D = F d

D/d = 4 so F/f = 4

If F=200 N f=?


MachinesMachines

Pulleys

Dd

f

F

Work in = Work out

f D = F d

D/d = 4 so F/f = 4

If F=200 N f = 200 N/ 4 = 50 N


MachinesMachines

Hydraulic machine

Dd

fF

Work in = Work out

f D = F d if D=20 cm , d =1 cm, and F= 800 N, what is the minimum force f?


MachinesMachines

Hydraulic machine

Dd

fF

Work in = Work out

f D = F d

f 20 cm = 800 N (1 cm) f = 40 N

if D=20 cm , d =1 cm, and F= 800 N, what is the minimum force f?


EfficiencyEfficiency

Effeciency Energyout

Energyin

Ein

Eout

Eloss


?in

out

Energy

EnergyEfficiency

Ein = 200 J

Eout= 150 J

Eloss= ?


?in

out

Energy

EnergyEfficiency

Ein = 200 J

Eout= 150 J

Eloss= 50J

=0.75=75%

Two Machines e1 and e2 Two Machines e1 and e2 connected to each other in seriesconnected to each other in series

Two Machines e1 and e2Two Machines e1 and e2

Eout=eff (Ein)=0.5(100J)=50J


Total efficiency when 2 machines are connected one after the other is etot=e1 (e2)

Kinetic Energy, KEKinetic Energy, KE

KE =1/2 m v2

m=2.0 kg and v= 5 m/sKE= ?

Kinetic EnergyKinetic Energy

KE =1/2 m v2m=2.0 kg and v= 5 m/sKE= 25 J




m=4.0 kg and v= 5 m/sKE= 50J



m=2.0 kg and v= 10 m/sKE= 100J

22

2

mv2

142vm

2

1KE

2v vif

mv2

1KE

Double speed and KE increases by 4


KE =1/2 m v2

if m doubles KE doubles if v doubles KE quadruples if v triples KE increases 9x if v quadruples KE increases ____ x

Work Energy TheormWork Energy Theorm

KE =1/2 m v2

F = m a


K =1/2 m v2

F = m a F d = m a d


KE =1/2 m v2

F = m a F d =m a d F d = m (v/t) [(v/2)t]


K E=1/2 m v2

F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v2


KE =1/2 m v2

F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v2

W = ∆ KE

Work EnergyWork Energy W = ∆KE

How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?


How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?

W= ∆KE =-1/2 m v2

=-1/2(2000 kg)(20 m/s)2

= - 1000kg (400 m 2 /s 2) = - 400,000 Joules


How much work is required to stop a 2000 kg car traveling at 20 m/s? If the friction force equals its weight, how far will it skid?

W= ∆K = - 400,000 Joules F=weight=mg=-20,000 N

W=F d d=W/F=-400,000 J/-20,000N = 20.0 m

Work EnergyWork Energy W = ∆KE v = 20 m/s

d=? m

v = 10 m/s

d= 15 m

Same Friction Force

Work EnergyWork Energy W = ∆KE v = 20 m/s

d=60m(4 times 15m)

v = 10 m/s

d= 15 m

Same Friction Force

Potential Energy, PEPotential Energy, PE

• Gravitational Potential Energy Gravitational Potential Energy • SpringsSprings• ChemicalChemical• PressurePressure• Mass (Nuclear)Mass (Nuclear)

• Measured in JoulesMeasured in Joules

Potential Energy, PEPotential Energy, PE

• Gravitational Potential Energy Gravitational Potential Energy • SpringsSprings• ChemicalChemical• PressurePressure• Mass (Nuclear)Mass (Nuclear)

The energy required to put something in its place (state)

Potential EnergyPotential Energy

Gravitational Potential Energy = weight x height

PE=(mg) h

4.0 m

m = 2.0 kg

Potential EnergyPotential EnergyPE=(mg) h

4.0 m

m = 2.0 kg

K=?

PE=80 J

Potential Energy to Kinetic EnergyPotential Energy to Kinetic EnergyPE=(mg) h

2.0 m

m = 2.0 kg

KE=?

PE=40 J

1.0 m

K E= 0 J

Conservation of EnergyConservation of Energy

Total Mechanical Energy, E = PE +K

Energy can neither be created nor destroyed only transformed from one form to another

In the absence of friction or other non-conservative forces the total mechanical energy of a system does not change

E f=Eo


10.0 m

m = 1.02 kg (mg = 10.0 N)

K = 0 JPE=100 J

PE = 75 J

PE = 50 J

PE = 0 J

PE= 25 J

K = ?

K= ?

K = 50 J

K = 25 J Constant E{E = K + PE}

Ef = Eo

No frictionNo Air resistance


5.0 m

m = 2.0 kgK=0 J

PE=100 J

PE = 0 J

K = ?

Constant E{E = K + U}Constant E{E = K + PE}Ef=Eo

No friction


5.0 m

m = 2.0 kgK = 0 J

PE =100 J

v = ?

K = 100 J

Constant E{E = K + U}Constant E{E = K + PE}Ef=Eo

No friction

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