Work and Energy Dr. Robert MacKay Clark College

Introduction What is Energy? What are some of the different forms of energy? Energy = $$$

Overview W KE PE 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 W KE PE Work Kinetic Energy Potential Energy Heat Loss

Crib Sheet

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 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?

Actual mechanical advantage is usually less.

https://youtu.be/4F7XqodHewk

Machines Pulleys f D d 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 Pulleys f Work in = Work out f D = F d D d F

Machines Pulleys f D d 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 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

Machines Pulleys f D d 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 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

Machines F f Hydraulic machine d D 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 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?

Machines F f Hydraulic machine d D f = 40 N Work in = Work out 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 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=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

Kinetic Energy KE =1/2 m v2 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

Double speed and KE increases by 4

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]

Work Energy Theorm 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

Work Energy Theorm 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 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 W= ∆KE =-1/2 m v2 =-1/2(2000 kg)(20 m/s)2 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 W= ∆K = - 400,000 Joules F=weight=mg=-20,000 N 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

Work Energy W = ∆KE d=60m (4 times 15m) v = 20 m/s Same Friction Force

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 PE=(mg) h Gravitational Potential Energy = weight x height PE=(mg) h m = 2.0 kg 4.0 m

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 = 0 J m = 1.02 kg (mg = 10.0 N) Constant E {E = K + PE} Ef = Eo PE = 75 J K = 25 J 10.0 m PE = 50 J K = 50 J PE= 25 J K= ? No friction No Air resistance PE = 0 J K = ?

Conservation of Energy PE=100 J m = 2.0 kg K=0 J Constant E {E = K + U} Constant E {E = K + PE} Ef=Eo 5.0 m No friction PE = 0 J K = ?

Conservation of Energy PE =100 J m = 2.0 kg K = 0 J Constant E {E = K + U} Constant E {E = K + PE} Ef=Eo 5.0 m No friction v = ? K = 100 J