Work and Energy Dr. Robert MacKay Clark College Introduction What is Energy? What are some of the different forms of energy? Energy = $$$ Overview Work (W) Kinetic Energy (KE) Potential Energy (PE) All Are measured in Units of Joules (J) 1.0 Joule = 1.0 N m W KE PE Overview Work Kinetic Energy Potential Energy Heat Loss W Heat Loss KE PE Heat Loss Crib Sheet W FD// (JOULES) 1 2 KE mv 2 GPE mgh 1 2 SPE kx F kx 2 W P (J / s Watt) t Wnet KE E KE PE E f E 0 W NC Work and Energy Work = Force x distance W=Fd Actually Work = Force x Distance parallel to force d=4.0 m F= 6.0 N W= F d = 6.0 N (4.0m) = 24.0 J Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W=? Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W = 80 J Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= - 6.0 N W= F d = -6.0 N (8.0m) =-48 J Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d =?J Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d = -30 J Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= ? N W= 60 J Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= 10 N W= 60 J Work and Energy Work = Force x Distance parallel to force d= ? m F= - 50.0 N W= 200 J Work and Energy Work = Force x Distance parallel to force d= -4.0 m F= - 50.0 N W= 200 J Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= + 6.0 N W= 0 (since F and d are perpendicular Power Work J / s Power time Work = Power x time 1 Watt= 1 J/s 1 J = 1 Watt x 1 sec 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J Energy = $$$$$$ 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=1m 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 Levers d=1m D =8 m 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 Work in = Work out f D = F d D f 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=? D f 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 D f 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 Hydraulic machine D F f 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 Hydraulic machine F f 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 Energy out Effeciency Energy in Eout Ein Eloss Efficiency Energyout Efficiency ? Energyin Eout= 150 J Ein = 200 J Eloss= ? Efficiency Energyout Efficiency ? =0.75=75% Energyin Eout= 150 J Ein = 200 J Eloss= 50J Two Machines e1 and e2 connected to each other in series Two Machines e1 and e2 Eout=eff (Ein)=0.5(100J)=50J Two Machines e1 and e2 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 KE =1/2 m v2 m=4.0 kg and v= 5 m/s KE= ? m=2.0 kg and v= 5 m/s KE= 25 J Kinetic Energy KE =1/2 m v2 m=4.0 kg and v= 5 m/s KE= 50J m=2.0 kg and v= 5 m/s KE= 25 J Kinetic Energy KE =1/2 m v2 m=2.0 kg and v= 5 m/s KE= 25 J m=2.0 kg and v= 10 m/s KE= ? Kinetic Energy KE =1/2 m v2 m=2.0 kg and v= 5 m/s KE= 25 J m=2.0 kg and v= 10 m/s KE= 100J 1 2 KE mv 2 if v 2v 1 1 2 2 KE m2v 4 mv 2 2 Double speed and KE increases by 4 Kinetic Energy KE if =1/2 m v2 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=ma Work Energy Theorm K =1/2 m v2 F=ma Fd=m a d Work Energy Theorm KE =1/2 m v2 F=ma F d =m a d F d = m (v/t) [(v/2)t] Work Energy Theorm K E=1/2 m v2 F=ma Fd=mad F d = m (v/t) [(v/2)t] W = 1/2 m v2 Work Energy Theorm KE =1/2 m v2 F=ma Fd=mad F d = m (v/t) [(v/2)t] W = 1/2 m v2 W = ∆ KE Work Energy W = ∆KE How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)? Work Energy W = ∆KE 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 Work Energy W = ∆KE 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 Energy W = ∆KE v = 20 m/s d=? m Same Friction Force v = 10 m/s d= 15 m Work Energy W = ∆KE v = 20 m/s d=60m (4 times 15m) Same Friction Force v = 10 m/s d= 15 m Potential Energy, PE • • • • • Gravitational Potential Energy Springs Chemical Pressure Mass (Nuclear) • Measured in Joules Potential Energy, PE The energy required to put something in its place (state) • • • • • Gravitational Potential Energy Springs Chemical Pressure Mass (Nuclear) Potential Energy Gravitational Potential Energy = weight x height PE=(mg) h 4.0 m m = 2.0 kg Potential Energy PE=(mg) h PE=80 J m = 2.0 kg 4.0 m K=? Potential Energy to Kinetic Energy PE=(mg) h m = 2.0 kg K E= 0 J PE=40 J 2.0 m 1.0 m KE=? Conservation of Energy Energy can neither be created nor destroyed only transformed from one form to another Total Mechanical Energy, E = PE +K In the absence of friction or other non-conservative forces the total mechanical energy of a system does not change E f=Eo Conservation of Energy PE=100 J K=0J m = 1.02 kg (mg = 10.0 N) PE = 75 J K = 25 J 10.0 m PE = 50 J K = 50 J PE= 25 J PE = 0 J Constant E {E = K + PE} Ef = Eo K= ? K=? No friction No Air resistance Conservation of Energy PE=100 J m = 2.0 kg K=0 J Constant E {E = K + PE} U} Ef=Eo 5.0 m No friction PE = 0 J K=? Conservation of Energy PE =100 J m = 2.0 kg K=0J Constant E {E = K + PE} U} Ef=Eo 5.0 m No friction v=? K = 100 J