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Learn about the concepts of work and energy, the different forms of energy, and how to calculate work and power. Understand the relationship between force, distance, and energy. Explore the use of machines and pulleys to increase force or distance. Examples and calculations are provided. Suitable for students and anyone interested in understanding the principles of 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.
How much force must be Ideally applied at the handle of the “wheel and axel arrangement of an old fashioned water well to lift a 60 N bucket of water. Assume the diameter of the inner log to be 10 cm and the radius of the handle to be 25 cm.
A force of 300 N in applied to the handle 30 cm from the vertical bar. How much digging for does this provide at the bottom blades assumed to be 3 cm from the vertical line?
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= ?