 # Work and Energy Dr. Robert MacKay Clark College.

## Presentation on theme: "Work and Energy Dr. Robert MacKay Clark College."— Presentation transcript:

Work and Energy Dr. Robert MacKay Clark College

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

Overview W K U Work (W) Kinetic Energy (K) Potential Energy (U)
All Are measured in Units of Joules (J) 1.0 Joule = 1.0 N m W K U

Overview W K U Work Kinetic Energy Potential Energy Heat Loss

Work and Energy Dr =4.0 m W= F Dr = 6.0 N (4.0m) = 24.0 J F= 6.0 N
Work = Force x displacement W = F d Actually Work = Force x displacement parallel to force Dr =4.0 m W= F Dr = 6.0 N (4.0m) = 24.0 J F= 6.0 N

Work and Energy Dr = 8.0 m F= - 6.0 N W= F Dr = -6.0 N (8.0m) =-48 J
Work = Force x Displacement parallel to force Dr = 8.0 m F= N W= F Dr = -6.0 N (8.0m) =-48 J

Work and Energy Dr = 6.0 m F= ? N W= 60 J
Work = Force x Displacement parallel to force Dr = 6.0 m F= ? N W= 60 J

Work and Energy Dr = ? m F= - 50.0 N W= 200 J
Work = Force x Displacement parallel to force Dr = ? m F= N W= 200 J

Work and Energy Dr = 8.0 m F= + 6.0 N W= 0
Work = Force x Displacement parallel to force Dr = 8.0 m F= N W= 0 (since F and d are perpendicular

Figure 7.2  If an object undergoes a displacement ∆r under the action of a constant force F, the work done by the force is F∆r cos . Fig. 7.2, p.184

Work and Energy Dr = 8.0 m F= + 6.0 N 40° Work = F Dr
Dot product W = F Dr cos (40) Dr = 8.0 m F= N 40° W= (6.0N) [8.0m cos(40) ]= 36.8 J

Work and Energy Dr = 8.0 m F= + 6.0 N 40° Work = F Dr
Dot product W = F Dr cos(40) Dr = 8.0 m F= N 40° W= (6.0N cos(40) ) [8.0m]= 36.8 J

Total Area ~S (DA) Work ~ S (FxDx) ~ ~ Fig. 7.7a, p.189
Figure 7.7  (a) The work done by the force component Fx for the small displacement ∆x is Fx ∆x, which equals the area of the shaded rectangle. The total work done for the displacement from xi to xf is approximately equal to the sum of the areas of all the rectangles. Fig. 7.7a, p.189

Figure 7.7  (b) The work done by the component Fx of the varying force as the particle moves from xi to xf is exactly equal to the area under this curve. Fig. 7.7b, p.189

Work Variable Force (q=0)
Work = F Dr cosq = Favg Dx

Springs 101 Spring Constant k, stiffness = 50 N/m

Work Variable Force (q=0)
Work = F D cosq = F D

Work Variable Force (q=0)
Work = “Area” units of N m (Joules) W= 0.5*(100N)(4m) - 0.5(50N)2m) = +200 J -50 J = 150 J

Potential Energy, U Gravitational Potential Energy Springs Chemical
Pressure Mass (Nuclear) Measured in Joules

Potential Energy, U The energy required to put
something in its place (state) Gravitational Potential Energy Springs Chemical Pressure Mass (Nuclear)

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

Potential Energy U=(mg) h U=80 J m = 2.0 kg 4.0 m K=?

Potential Energy to Kinetic Energy
U=(mg) h m = 2.0 kg K E= 0 J PE=40 J 2.0 m 1.0 m KE=?

Potential Energy of a spring
x 1 U= kx2 2

Potential Energy of a spring
x 1 U= kx2 2 For a spring with stiffness k= 80 N/m, what is its potential energy when stretched 0.1m? How about 0.2 m?

Potential Energy of a spring
x 1 1 U= kx2 = 80 N/m (0.1m)2 2 2 = 0.40 J For a spring with stiffness k= 80 N/m, what is its potential energy when stretched 0.1m? How about 0.2 m?

Potential Energy of a spring
x 1 1 U= kx2 = 80 N/m (0.2m)2 2 2 = 1.60 J For a spring with stiffness k= 80 N/m, what is its potential energy when stretched 0.1m? How about 0.2 m?

Potential Energy of a spring
x 1 U= kx2 2

Kinetic Energy K=1/2mv2 Table 7.1, p.194

Kinetic Energy, K K =1/2 m v2 m=2.0 kg and v= 5 m/s K= ?

Kinetic Energy K =1/2 m v2 m=2.0 kg and v= 5 m/s K= 25 J

Kinetic Energy K =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 K =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 K =1/2 m v2 F = m a F d =m a d
F d = m (v/t) [(v/2)t]

Work Energy Theorm K =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 = ∆K How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?

Work Energy W = ∆K W= ∆K =-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= ∆K =-1/2 m v2 =-1/2(2000 kg)(20 m/s)2 = kg (400 m 2 /s 2) = - 400,000 Joules

Work Energy W = ∆K 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 = ∆K v = 20 m/s d=? m Same Friction Force v = 10 m/s

Conservation of Energy
Energy can neither be created nor destroyed only transformed from one form to another Total Mechanical Energy, E = U +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
U=100 J K = 0 J m = 1.02 kg (mg = 10.0 N) Constant E {E = K + U} Ef = Eo U = 75 J K = 25 J 10.0 m U = 50 J K = 50 J U = 25 J K= ? No friction No Air resistance U = 0 J K = ?

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

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

Conservation of Energy
Constant E {E = K + U} Ef=Eo Ef=Eo+Wother Kf +Uf=Ko+Uo +Wother

Example 8.8

L cos(25°)

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 = \$0.04 = 4 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 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
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 E =P t [ kW-hr=(kW) hr] Energy = ?
when 2000 watts (2 kW) of power are delivered for 6.0 hr. Cost at 4 cent per kW-hr?

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

Machines d = 1 m D =8 m Levers f=10 N F=? Work in = Work out f D = F d

Machines d = 1 m D =8 m Levers f=10 N F=? Work in = Work out
10N 8m = F 1m F = 80 N

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
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
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
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
f D = F d f 20 cm = 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= ?

Machines F f Hydraulic machine d D f = 40 N Work in = Work out
f D = F d f 20 cm = N (1 cm) f = 40 N if D=20 cm , d =1 cm, and F= 800 N, what is the minimum force f?

Machines F f Hydraulic machine d D if D=20 cm , d =1 cm, F= 800 N,
and f=60 N, what is the effeciency? Work in = 60 N (20 cm)=1200 N-cm Work out = 800 N (1 cm)= 800 N-cm