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Work and EnergyWork and Energy
PhysicsPhysics
Mr. DayMr. Day
Work WorkWork - the product of the magnitudes of - the product of the magnitudes of
the component of a force along the the component of a force along the direction of displacement and the direction of displacement and the displacementdisplacement
W = F dW = F d WorkWork
– Push a chair from rest to a velocityPush a chair from rest to a velocity Not WorkNot Work
– Hold a book in the airHold a book in the air– Carry a chair across the room at a constant Carry a chair across the room at a constant
velocityvelocity
Work (cont.)
Work is only done when the Work is only done when the components of a force are parallel to a components of a force are parallel to a displacementdisplacement
d
F
F
d
W = F d•All of the force is acting on the box
W = F d cos •Only the horizontal component of the force does work
Sign Convention for Work
Visual Concept
Work (cont.)
• Examples
d
F
d
F
d
F
W = F d cos
W = F d cosW = F d cos 0
W= F d
W = F d cosW = F d cos 90
W = 0
d F
W = F d cos
Work (cont.)
UnitsUnits– F d = WF d = W– N m = Joule (J)N m = Joule (J)
Work is a scalar quantityWork is a scalar quantity– Work can be positive or negativeWork can be positive or negative
• Positive Work - Force and displacement are in the Positive Work - Force and displacement are in the same directionsame direction
• Negative Work - Force and displacement are in Negative Work - Force and displacement are in opposite directionsopposite directions
Work Defined
Video
EX: A 20.0 kg suitcase is raised 3.0 m above a platform by a conveyor belt. How much work is done on the suitcase
m = 20.0 kgm = 20.0 kgd = 3.0 md = 3.0 m
W = F dW = F dW = (m a) dW = (m a) d
W = (20.0 kg)(9.8 m/sW = (20.0 kg)(9.8 m/s22)(3.0 m))(3.0 m)
W = 588 JW = 588 J
EX: A person drags a suitcase with a 100.0 N force at an angle of 60.0° for 200.0 m. How much work does she do?
F = 100.0 NF = 100.0 N
= 60.0°= 60.0°
d = 200.0 md = 200.0 m
d
F
W = F d cos W = F d cos W= (100.0 N)(200.0 m) cos 60°W= (100.0 N)(200.0 m) cos 60°
W = 10,000 JW = 10,000 J
W = 1.0 X 10W = 1.0 X 1044 J J
Types of Energy
Video
Energy
Kinetic EnergyKinetic Energy - the energy of an object - the energy of an object due to its motiondue to its motion– Depends on speed and massDepends on speed and mass– KE = 1/2 m vKE = 1/2 m v22
– Units Units Joules (J) Joules (J) Potential EnergyPotential Energy - energy associated - energy associated
with an object due to its positionwith an object due to its position– Units Units Joules (J) Joules (J)
Kinetic and Potential Energy
Video
Kinetic Energy
Visual Concept
EX: A 6.0 kg cat runs after a mouse at 10.0 m/s. What is the cat’s kinetic energy?
mmcc = 6.0 kg = 6.0 kg
vvcc = 10.0 m/s = 10.0 m/s
KEKEcc = 1/2 m = 1/2 mcc v vcc22
KEKEcc = 1/2 (6.0 kg)(10.0 m/s) = 1/2 (6.0 kg)(10.0 m/s)22
KEKEcc = 300 J = 300 J
EX: If a .10 kg mouse runs as fast as the cat, what is its kinetic energy?
mmmm = .10 kg = .10 kg
vvmm = 10.0 m/s = 10.0 m/s
KEKEmm = 1/2 m = 1/2 mmm v vmm22
KEKEmm = 1/2 (.10 kg)(10.0 m/s) = 1/2 (.10 kg)(10.0 m/s)22
KEKEmm = 5 J = 5 J
EX: A 24 kg dog begins to chase the cat and has the same kinetic energy as the cat. What is the dog’s velocity?
mmdd = 24 kg = 24 kg
KEKEdd = 300 J = 300 J
KEKEdd = 1/2 m = 1/2 mdd v vdd22
vvdd = √(2 Ke = √(2 Kedd / m / mdd))
vvdd = √(2 (300 J) / 24 kg) = √(2 (300 J) / 24 kg)
vvdd = 5.0 m/s = 5.0 m/s
Potential Energy Gravitational Potential EnergyGravitational Potential Energy - -
potential energy associated with an potential energy associated with an object due to its position relative to object due to its position relative to Earth or some other gravitational sourceEarth or some other gravitational source
PEPEgg = m g h = m g h
Elastic Potential EnergyElastic Potential Energy - the potential - the potential energy in a stretched or compressed energy in a stretched or compressed elastic objectelastic object– SpringSpring– Rubber bandRubber band
Potential Energy
Visual Concept
Elastic Potential Energy The length of a spring when no external The length of a spring when no external
forces are acting on it is called the relaxed forces are acting on it is called the relaxed lengthlength
PEPEee = 1/2 k x = 1/2 k x22
– PEPEee = 1/2 (spring constant)(distance stretched = 1/2 (spring constant)(distance stretched
or compressed)or compressed)22
Spring constantSpring constant - a parameter that - a parameter that expresses how resistant a spring is to being expresses how resistant a spring is to being compressed or stretchedcompressed or stretched
Elastic Potential Energy
Spring Constant
High spring constant High spring constant stiff spring stiff spring Low spring constant Low spring constant flexible spring flexible spring k = F / dk = F / d
– Units Units N / m N / m
Spring Constant
Visual Concept
EX: When a 2.00 kg mass is attached to a vertical spring, the spring is stretched 10.0 cm so the mass is 50.0 cm above the table.
m = 2.00 kgx = 10. cm = .10 mh = 50.0 cm = .50 m
A. What is the gravitational potential energy associated with the mass relative to the table?
PEPEgg = mgh = mgh
PEPEgg = (2.00 kg)(9.8 m/s = (2.00 kg)(9.8 m/s22)(.50 m))(.50 m)
m = 2.00 kgx = 10. cm = .10 mh = 50.0 cm = .50 m
PEPEgg = 9.8 J = 9.8 J
B. What is the elastic potential energy if the spring constant is 400.0 N/m?
PEPEee = 1/2 k x = 1/2 k x22
PEPEee = 1/2 (400.0 N/m)(.10m) = 1/2 (400.0 N/m)(.10m)22
m = 2.00 kgx = 10. cm = .10 mh = 50.0 cm = .50 mk = 400.0 N/m
PEPEee = 2.00 J = 2.00 J
C. What is the total potential energy of the system?
∑∑PE = PEPE = PEgg + PE + PEee
∑∑PE = 9.8 J + 2.00 JPE = 9.8 J + 2.00 J
m = 2.00 kgx = 10. cm = .10 mh = 50.0 cm = .50 mk = 400.0 N/m
PEPEee = 11.8 J = 11.8 J
Mechanical Energy
There are many types of energy There are many types of energy associated with a systemassociated with a system– KineticKinetic– Gravitational potentialGravitational potential– Elastic potentialElastic potential– ChemicalChemical– ThermalThermal
• Most can be ignored because they are Most can be ignored because they are negligible or not relevantnegligible or not relevant
Mechanical Energy
Mechanical energyMechanical energy - the sum of the - the sum of the kinetic and all forms of potential energykinetic and all forms of potential energy
ME = ∑KE + ∑PEME = ∑KE + ∑PE All other forms of energy are classified All other forms of energy are classified
as non-mechanical energyas non-mechanical energy
Conservation of Energy
Conserve means it remains the sameConserve means it remains the same Conservation of mechanical energyConservation of mechanical energy
MEMEii = ME = MEff
∑∑KEKEi i + ∑PE+ ∑PEgigi + ∑PE + ∑PEeiei = ∑KE = ∑KEf f + ∑PE+ ∑PEgfgf + ∑PE + ∑PEefef
In the presence of friction, energy is “lost” to In the presence of friction, energy is “lost” to heat energyheat energy
Niagara Falls and Energy Transformation
Video
Conservation of Mechanical Energy
Visual Concept
As the roller coaster falls the energy is transformed As the roller coaster falls the energy is transformed from potential energy to kinetic energyfrom potential energy to kinetic energy
The energy is then transferred back into potential The energy is then transferred back into potential energy, etc.energy, etc.
Energy of a roller coaster
Energy is a sling shot It starts with elastic potential energyIt starts with elastic potential energy It quickly transfers into kinetic energyIt quickly transfers into kinetic energy As the height increases it transfers into gravitational As the height increases it transfers into gravitational
energyenergy As it falls the energy transfers into kinetic energyAs it falls the energy transfers into kinetic energy
EX: A small 10.0 g ball is held to a slingshot that is stretched 6.0 cm. The spring constant of the band on the slingshot is 2.0 X 102 N/m.A. What is the elastic potential energy of the slingshot before it is released
PEPEee = 1/2 k x = 1/2 k x22
PEPEee = 1/2 (2.0 X 10 = 1/2 (2.0 X 102 2 N/m)(.06 m)N/m)(.06 m)22
m = 10.0 g = .0100 kgx = 6.0 cm = .06 mk = 2.0 X 102 N/m
PEPEee = .36 J = .36 J
B. What is the kinetic energy of the ball just after the slingshot is released?
MEMEii = ME = MEff
.36 J = ∑KE.36 J = ∑KEff
∑∑KEKEi i + ∑PE+ ∑PEgigi + ∑PE + ∑PEeiei = ∑KE = ∑KEf f + ∑PE+ ∑PEgfgf + ∑PE + ∑PEefef
∑∑PEPEeiei = ∑KE = ∑KEf f
C. What is the balls speed at the instant it leaves the slingshot?
KEKEf f = 1/2 m v= 1/2 m v22
v = √ (2KEv = √ (2KEff / m) / m)
v = √(2(.36 J) / (.01 kg))v = √(2(.36 J) / (.01 kg))
v = 8.5 m/sv = 8.5 m/s
D. How high would the ball travel if it were shot directly upward?
MEMEii = ME = MEff
.36 J = ∑PE.36 J = ∑PEff
∑∑KEKEi i + ∑PE+ ∑PEgigi + ∑PE + ∑PEeiei = ∑KE = ∑KEf f + ∑PE+ ∑PEgfgf + ∑PE + ∑PEefef
∑∑KEKEii = ∑PE = ∑PEgfgf
PEPEff = mgh = mgh
h = PEh = PEff / mg / mg
h = .36 J / ((.01 kg)(9.8 m/sh = .36 J / ((.01 kg)(9.8 m/s22))))h = 3.7 mh = 3.7 m
Work and Energy
Work-kinetic energy theoremWork-kinetic energy theorem - the net - the net work done on an object is equal to the work done on an object is equal to the change in the kinetic energy of the change in the kinetic energy of the objectobject
WWnetnet = ∆ KE = ∆ KE
WWfrictionfriction = ∆ ME = ∆ ME
Work Kinetic Energy Theorem
Visual Concept
Potential Energy is transferred into Kinetic EnergyPotential Energy is transferred into Kinetic Energy Next the change in the Kinetic Energy is equal to the Next the change in the Kinetic Energy is equal to the
net worknet work
Stopping Distance
If an object If an object has a has a higher higher kinetic kinetic energy, energy, more work more work is required is required to stop the to stop the objectobject
Ex: On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does it travel if the coefficient of kinetic friction between the sled and the ice is .10?
m = 10.0 kgm = 10.0 kg
vvii = 2.2 m/s = 2.2 m/s
vvff = 0 m/s = 0 m/s
µµkk = .10 = .10
vi W = ∆KEW = ∆KE
F d = KEF d = KEff - KE - KEii
µµkk = F = Fkk / F / FNN
FFkk = µ = µkk (-mg) (-mg)µµkk (-mg) d = 1/2 m v (-mg) d = 1/2 m vff
22 - 1/2 m v - 1/2 m vii22
µµkk (-mg) d = - 1/2 m v (-mg) d = - 1/2 m vii22
d = (- 1/2 m vd = (- 1/2 m vii22) / ) / µµkk (-mg) (-mg)
d = (- 1/2 (10.0 kg) (2.2 m/s)d = (- 1/2 (10.0 kg) (2.2 m/s)22) / ) /
(.10 (-10.0 kg) (9.8 m/s(.10 (-10.0 kg) (9.8 m/s22))))d = 2.47 md = 2.47 m
EX: A 10.0 kg shopping cart is pushed from rest by a 250.0 N force against a 50.0 N friction force over 10.0 m distance.
m = 10.0 kgm = 10.0 kg
vvii = 0 = 0
FFpp = 250.0 N = 250.0 N
FFkk = 50.0 N = 50.0 N
d = 10.0 m d = 10.0 m
FpFk
FN
Fg
A. How much work is done by each force on the cart?
WWpp = F = Fpp d cos d cos
WWpp = (250.0 N)(10.0 m)cos 0 = (250.0 N)(10.0 m)cos 0WWpp = 2500 J = 2500 J
WWkk = F = Fkk d cos d cos
WWkk = (50.0 N)(10.0 m)cos 180 = (50.0 N)(10.0 m)cos 180WWkk = -500 J = -500 J
WWgg = 0 = 0
WWNN = 0 = 0
B. How much kinetic energy has the cart gained? WWnetnet = ∆KE = ∆KE
WWpp + W + Wkk = KE = KEff - KE - KEii
2500 J + -500 J = KE2500 J + -500 J = KEff - 0 - 0KEKEff = 2000J = 2000J
C. What is the carts final speed?KE = 1/2 m vKE = 1/2 m v22
v = √((2KE)/(m))v = √((2KE)/(m))v = √((2(2000 J))/(10.0 kg))v = √((2(2000 J))/(10.0 kg))
v = 20 m/sv = 20 m/s
Power PowerPower - the rate at which energy is - the rate at which energy is
transferredtransferred P = W / ∆ tP = W / ∆ t
– Units Units J / s J / s watt (w) watt (w) Since P = W / ∆ t and W = F d;Since P = W / ∆ t and W = F d;
– P = F d / ∆ t P = F d / ∆ t P = F (d / t) P = F (d / t)– P = F vP = F v
Horsepower is another unit of powerHorsepower is another unit of power– 1 hp = 746 w1 hp = 746 w
Power Defined
Video
Power
Visual Concept
EX: A 100.0 N force moves an object 20.0 m in 5.0 s. What is the power?
F = 100.0 NF = 100.0 N
d = 20.0 md = 20.0 m
t = 5.0 st = 5.0 s
P = 400 wP = 400 w
P = F (d / t)P = F (d / t)
P = 100.0 N (20.0 m / 5.0 s)P = 100.0 N (20.0 m / 5.0 s)
d = 20.0 m
F = 100.0 N
EX: Two horses pull a cart. Each exerts a 250 N force at a 2.0 m/s speed for 10.0 min.FFh1h1 = 250 N = 250 N
FFh2h2 = 250 N = 250 N
v = 2.0 m/sv = 2.0 m/s
∆ ∆ t = 10.0 min t = 10.0 min
= 600 = 600 ss
v = 2.0 m/s
F = 250 N
F = 250 N
A. Calculate the power delivered by the forces.
PPh1h1 = F = Fh1h1 v v
PPh1h1 = (250 N)(2.0 m/s) = (250 N)(2.0 m/s)PPh1h1 = 500 w = 500 w PPh2h2 = F = Fh2h2 v v
PPh2h2 = (250 N)(2.0 m/s) = (250 N)(2.0 m/s)PPh2h2 = 500 w = 500 w
∑∑P = PP = Ph1h1 + P + Ph2h2
∑∑P = 500 w + 500wP = 500 w + 500w∑∑P = 1000 wP = 1000 w
B. How much work is done by the two horses?
P = W / ∆ tP = W / ∆ t
W = P ∆ tW = P ∆ t
W = (1000 w)(600 s)W = (1000 w)(600 s)W = 6.0 X 10W = 6.0 X 1055 J J
Work Cited
SourcesSources– www.classroomphysics.comwww.classroomphysics.com– www.clipart.comwww.clipart.com– Holt PhysicsHolt Physics– United StreamingUnited Streaming