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