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## Work and Energy

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**Work and Energy**Dr. Robert MacKay Clark College**Introduction**• What is Energy? • What are some of the different forms of energy? • Energy = $$$**Overview**• Work (W) Kinetic Energy (KE) Potential Energy (PE) • All Are measured in Units of Joules (J) • 1.0 Joule = 1.0 N m W KE PE**Overview**• Work Kinetic Energy Potential Energy Heat Loss W KE PE Heat Loss Heat Loss**Work and Energy**• Work = Force x distance • W = F d • Actually • Work = Force x Distance parallel to force d=4.0 m W= F d = 6.0 N (4.0m) = 24.0 J F= 6.0 N**Work and Energy**• Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W = ?**Work and Energy**• Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W = 80 J**Work and Energy**• Work = Force x Distance parallel to force d= 8.0 m F= - 6.0 N W= F d = -6.0 N (8.0m) =-48 J**Work and Energy**• Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d = ? J**Work and Energy**• Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d = -30 J**Work and Energy**• Work = Force x Distance parallel to force d= 6.0 m F= ? N W= 60 J**Work and Energy**• Work = Force x Distance parallel to force d= 6.0 m F= 10 N W= 60 J**Work and Energy**• Work = Force x Distance parallel to force d= ? m F= - 50.0 N W= 200 J**Work and Energy**• Work = Force x Distance parallel to force d= -4.0 m F= - 50.0 N W= 200 J**Work and Energy**• Work = Force x Distance parallel to force d= 8.0 m F= + 6.0 N W= 0 (since F and d are perpendicular**Power**• Work = Power x time • 1 Watt= 1 J/s • 1 J = 1 Watt x 1 sec • 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J • Energy = $$$$$$ • 1 kW-hr = $0.08 = 8 cents**Power**• 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.**Power**• 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.**Power**• Energy = Power x time • E =P t • [ kW-hr=(kW) hr] • or • [ J=(J/s) s= Watt * sec ]**Power**• 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, & s or kW-hr, kW, hr**Power**• 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, & s or kW-hr, kW, hr Energy=0.1 kWatt (24 hrs)=2.4 kWatt-hr**Power**• Energy = Power x time What is the power output of a duck who does 3000 J of work in 0.5 sec? What units should we use? J,W, & s or kW-hr, kW, hr**Power**• Energy = Power x time 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 What units should we use? J,W, & s or kW-hr, kW, hr**Power**• 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?**Power**• 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**Machines**d = 1 m D =8 m • Levers f=10 N F=? 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.**Machines**d = 1 m D =8 m • Levers f=10 N F=? Work in = Work out 10N 8m = F 1m F = 80 N 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.**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. Machines • Pulleys f Work in = Work out f D= F d D d F**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. Machines • Pulleys Work in = Work out f D= F d D/d = 4 so F/f = 4 If F=200 N f=? f D d F**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. Machines • Pulleys 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 f D d F**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. Machines F f • Hydraulic machine d D 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?**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. Machines F f • Hydraulic machine d D 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?**Efficiency**Eout Ein Eloss**Efficiency**Eout= 150 J Ein= 200 J Eloss= ?**Efficiency**=0.75=75% Eout= 150 J Ein= 200 J Eloss= 50J**Two Machines e1 and e2**Eout=eff (Ein)=0.5(100J)=50J**Two Machines e1 and e2**Total efficiency when 2 machines are connected one after the other is etot=e1 (e2)**Kinetic Energy, KE**• KE =1/2 m v2 m=2.0 kg and v= 5 m/s KE= ?**Kinetic Energy**m=2.0 kg and v= 5 m/s KE= 25 J • KE =1/2 m v2 m=4.0 kg and v= 5 m/s KE= ?**Kinetic Energy**m=2.0 kg and v= 5 m/s KE= 25 J • KE =1/2 m v2 m=4.0 kg and v= 5 m/s KE= 50J**Kinetic Energy**m=2.0 kg and v= 5 m/s KE= 25 J • KE =1/2 m v2 m=2.0 kg and v= 10 m/s KE= ?**Kinetic Energy**m=2.0 kg and v= 5 m/s KE= 25 J • KE =1/2 m v2 m=2.0 kg and v= 10 m/s KE= 100J**Kinetic Energy**• 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 Theorm**• KE =1/2 m v2 • F = m a**Work Energy Theorm**• K =1/2 m v2 • F = m a • F d = m a d**Work Energy Theorm**• KE =1/2 m v2 • F = m a • F d =m a d • F d = m (v/t) [(v/2)t]