Copyright Sautter 2003

WORK & ENERGY• Work, in a physics sense, has a precise definition,

unlike the common use of the word. When you do your home “work” you probably, from a physics stand point, did no work at all !

• Work is defined as force applied in the direction of the motion multiplied times the distance moved.

• When work is done by moving an object in a horizontal direction, work equals the applied force times the cosine of the angle of the applied force times the distance the object is moved.

• W = F (cos ) x s, (s stands for distance)• Work is a scalar quantity (it has no direction). The

sign of a work quantity (positive or negative) indicates the direction of energy flow as into or out of a system but does not give it a direction as in a vector quantity.

WORK & ENERGY• The terms work and energy are interchangeable.

Energy is defined as the ability to do work.• Kinds of work and energy• (1) mechanical work – work done by applying a

force over a distance• (2) work of friction – work required to overcome

friction• (3) gravitational potential energy – energy needed to

lift an object against the force of gravity• (4) elastic potential energy – the energy stored in a

compressed or stretched spring• (5) kinetic energy – energy an object has because of

its motion (velocity)

Distance moved

Appliedforce

Verticalcomponent

Horizontalcomponent

Vertical component = Applied force x sinHorizontal component = Applied force x cos

Work = force in direction of motion x distance moved

The horizontal force component is in the direction of the motion

FORCE OF FRICTION = 0WORK DONE = 0

FORCE OF FRICTION > 0WORK DONE = FFRICTION x DISTANCE

Recall: Ffriction = coefficient of friction x Fnormal

and on a horizontal surface:

Fnormal = weight of object = mass x gravity

GRAVITATIONAL POTENTIAL ENERGY

• When an object is lifted, work is done against the force of gravity (the weight of the object).

• Since weight is a force and the height to which an object is lifted is a distance, then force times distance equals work done.

• Weight of an object can be calculated using mass time gravity. When objects are lifted near the surface of the earth, gravity is assumed to be constant at 9.8 m/s2 (32 ft/s2).

• If object are lifted well beyond the earth’s surface gravity diminishes to progressively smaller values and the work done in the lifting becomes less and less.

GRAVITATIONAL POTENTIAL ENERGY

• Potential energy change equals weight times change in height.

• Weight equals mass times gravity• Potential energy change equals mass times

gravity times height (distance lifted)

MEASUREMENT OF POTENTIAL ENERGY IS RELATIVE

Boy Two!You’re at a

High potential

I sure am !

Who are they kidding ??

One

Two

Three

Two is at a higher potential energy than One but lower thanThree. Two’s potential energy is negative relative to Three’s and

positive relative to One’s.

If this point was used as reference,One, Two and Three would all have

negative potential energies.

Radius of Earth = 4000 miles

scale150 lbs

Two Radius of Earth = 8000 miles

scale37.5 lbs

Three Radius of Earth = 12000 miles

scale16.7 lbs

¼ wt

1/9 wt

Normalwt

Calculating Work in Different Gravitational Fields

• Potential energy changes are different in different gravitational field because the value of g changes.

• As seen in the previous slide, at an altitude of one earth radii above the earth (4000 miles) gravity is ¼ of normal gravity (1/4 x 9.8 m/s2 = 2.45 m/s2). At two earth radii altitude, gravity is 1.09 m/s2.

• An object of mass 10 kg is lifted 5 meters on earth. The work done (potential energy increase) is (P.E. = mgh) 10 kg x 9.8 m/s2 x 5 m = 490 joules.

• At one earth radii, work done is 10 kg x 2.45 m/s2 x 5 m = 122.5 joules (1/4 of the work done in lifting the same object on earth)

• At two earth radii above the earth (8000 miles altitude) the work done on the same object is 10 kg x 1.09 m/s2 x 5 m = 54.4 joules (1/9 of the work required to lift the object on earth)

KINETIC ENERGY• Kinetic energy is the energy of motion. In order to possess

kinetic energy an object must be moving.

• As the speed (velocity) of an object increases its kinetic energy increases. The kinetic energy content of a body is also related to its mass. The most massive objects at the same speed contain the most kinetic energy.

• Work = force x distance (W = F x s )

• Recall that F = mass x acceleration (F = m x a)

• Therefore: Work = m x a x s

• Also, for an object initially at rest, recall that acceleration equals the final velocity squared divided by twice the distance traveled: a = v2 / (2 s)

• Work = m (v2 / (2 s)) s, canceling out the distance term (s) gives, Work = (m v2 ) / 2 or 1/2 m v2

• Since the object is in motion, the work content is called kinetic energy and therefore: K.E. = 1/2 m v2

High kinetic energy.High velocity !

Kinetic energy = 0No motion !

ELASTIC POTENTIAL ENERGY

• Elastic potential energy refers to the energy which is stored in stretched of compressed items such as springs or rubber bands.

• The elongation or compression of elastic bodies is described by Hooke’s Law. This law relates the force applied to the elongation or compression experienced by the body.

• In plain words, Hooke’s Law says, “the harder you pull on a spring, the more it stretches”. This relationship is given by the equation: F = k x.

• F is the applied force, k is a constant called the spring constant or Hooke’s constant and x is the elongation of the spring.

• Springs with large k values are hard to stretch or compress such as a car spring. Those with small constants are easy of stretch or compress such as a slinky spring.

400grams

200grams

FORCE

(N)

ELONGATION (M)

Slope = spring constant600grams

\

Elongation of spring

Force

(N)

Distance (M) x

Constant forceWork = force x distance

Constant force

Distance moved

Force x distance equalsarea under the graph

Work = area under aforce versus distance graph

FORCE

(N)

ELONGATION (M)

X1 X2

F1

F1

Area under the graphgives the work to stretchthe spring. Work neededto stretch the spring to x2

is ½ F2 times x2

Work needed to stretch the spring to x1

is ½ F1 times x1

Work needed to stretch the spring from x1 to x2 is

(½ F2 times x2) – ( ½ F1 times x1)

Since F = kx and W = ½ Fx, W = ½ (kx) x orW = ½ kx2 and work from

x1 to x2 is given by:

W = ½ k (x22 – x1

2)

FORCE

(N)

DISPLACEMENT (M)

X1 X2

WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)

WIDTH OF EACH BOX = X

AREA MISSED - INCREASINGTHE NUMBER BOXES WILL

REDUCE THIS ERROR!

AS THE NUMBER OF BOXESINCREASES, THE ERROR

DECREASES!

BOX METHOD

Finding Area Under Curves Mathematically

• Areas under force versus distance graphs (work) can be found mathematically. The process requires that the equation for the graph be known and integral calculus be used.

• Recall that integration is also referred to as finding the antiderivative of a function.

• The next slide reviews the steps in finding the integral of the basic function, y = kxn.

INTEGRATION – THE ANTIDERIVATIVEINTEGRATION IS THE REVERSE PROCESS OF

FINDING THE DERIVATIVE. IT CAN ALSO BE USEDTO FIND THE AREA UNDER A CURVE.

THE GENERAL FORMAT FOR FINDING THE INTEGRAL OF A SIMPLE POWER RELATIONSHIP, Y = KXn

ADD ONE TO THE POWER

DIVIDE THEEQUATION

BY THE N + 1

ADD A CONSTANT

is the symbolfor integration

APPLYING THE INTEGRAL FORMULA

GIVEN THEEQUATION

FORMAT TO FIND THE INTEGRAL

Integration can be used to find area under a curve betweentwo points. Also, if the original equation is a derivate, then

the equation from which the derivate came can be determined.

APPLYING THE INTEGRAL FORMULAFind the area between x = 2 and x = 5 for the equation y = 5X3.First find the integral of the equation as shown on the previous

frame. The integral was found to be 5/4 X4 + C.

The values 5 and 2 arecalled the limits.

each of the limits isplaced in the integratedequation and the resultsof each calculation aresubtracted (lower limit

from upper limit)

Finding the Equation for Work Stored in Spring using Integration

Hooke’s Law F = kx

Work = Fdx = kx dxWork = k x1+1/ (1+1) = k x2 / 2 + C

W = ½ k x2

Note that the work equation is the same as that found by area under the curve

methods used in previous slides

x2

x1

|

Conservation of Energy• A most fundamental law of physics is the “Law of

Conservation of Energy”. It is the basis upon which much of physics is built.

• “Energy (ability to do work) cannot be created or destroyed, only changed in form”. This means that heat can be converted of electricity, electricity can be converted to motion, motion can be converted to heat, etc. In each and every case, all energy is conserved and can be accounted for as equal before and after the process.

• More fundamentally, potential energy can be converted to kinetic energy, kinetic energy to work of friction, elastic energy to kinetic energy, etc. all without net energy loss.

Conservation of Energy“The energy content of the universe is constant”

Energy change for a falling stone

Stone initially at rest (height is greatest velocity is zero).Potential energy is maximum. Kinetic energy is zero.

ground

Stone is half way to the ground. Potential energy is 1/2 maximum. Kinetic energy is 1/2 maximum .

Stone is just about to hit the ground. Potential energy is zero. Kinetic energy is maximum .

All potential energy has been converted to kinetic energy.

High potential energyLow kinetic energy

Low potential energyHigh kinetic energy

h KE = ½ mv2

PE = mg h

Power• Power is work divided by the time required to perform

the work. If two different energy sources do the same quantity of work, the one requiring the least time is the more powerful.Power can be measured in watts (joules / second) or horsepower (550 ft lbs / second).

• Power = work / time• Work = force x distance• Power = (force x distance) / time• Since distance divided by time equals velocity• Power = force x velocity

Work & Energy ProblemsA horizontal force pulls a box 5 meters across a floor with a force

of 420 N. The box weighs 500 N. How much work is done ?

500 N420 N

5 m

Work = F cos x sSince the weight of the boxis not the applied force it is

not related to the work done.

Horizontal means that = 00, cos 00 = 1.0W = 420 N x 1.0 x 5 m = 2100 joules

(Recall : 1 j = 1 N x 1 m)or 2.1 kilojoules

Work & Energy ProblemsA 60 kg box is pushed across a floor with a coefficient of sliding

friction equaling 0.30. If the box moves 12 meters at constant speed, how much work is done ?

60 kg? N

12 m

Ff

w = mgw = 60 x 9.8 = 588 N

Constant speed means noacceleration therefore the

net force must be zero.The applied force must equal

the force of friction

Recall: Ffriction = coefficient of friction x Fnormal

and on a horizontal surface:Fnormal = weight of object = mass x gravity

Ff = 0.30 x 588 N = 176 NWork = F cos x s, = 00

W = 176 N x (1.0) x 12 = 2112 j

Work & Energy ProblemsHow much work is needed to lift a 100 lb barbell from the

floor 1.5 feet over the head of a 5ft 6 inch man?

5’6’’

1.5’

7.0 ft

Lifting always involveschanging gravitational

potential energy.P.E. = mgh

weight is m x g and is expressed in pounds.

P.E. = mghP.E. = 100 lbs x 7 ft = 700 ft-lbs

Work & Energy ProblemsA 20.0 kg crate is pulled 8.00 meters up a frictionless incline with a 200 angle. How much work is done ?

The crate is actually beinglifted against gravity.

Although it is pulled 8 mit is lifted only the vertical

distance h.200

8 m

h

h = 8 x sin 200 = 2.74 mP.E. = mgh

P.E. = 20 kg x 9.8 m/s2 x 2.74 mP.E. = 536 j

How much work is usedIf the force of friction

Against the crate is 10.0 N ?

The crate is moved 8.00 meters against friction forces.

Work of friction = Ff x s, Wf = 10.0 N x 8.00 m = 80 jWork total = P.E. + Wf

Work total = 536 j + 80 j = 616 j

Work & Energy ProblemsA box slides down a frictionless incline 12 feet long with

an angle of 300. What is its speed at the bottom?

300

12 ft

h

The energy in the box is being converted from potential (it iselevated) to kinetic as it slides.Conservation of energy tells us

that all potential becomes kinetic.

P.E.box = K.E.box

mgh = ½ mv2 , since mass appears on both sides it can be divided out leaving:gh = ½ v2, rearranging the equation gives:

v = (2 x g x h)1/2

h is the vertical height = 12 sin 300 = 12 x 0.5 = 6 ftv = (2 x 32 ft/s2 x 6 ft)1/2 = 19.6 ft/s

Work & Energy ProblemsA 2.0 slug crate is pushed with a force of 100 lbs, 10 ft up an incline of 300

which has a coefficient of friction of 0.10. Find the speed of the crate.

300

10 ft

h100 lbs

Conservation of energy tellsus that all work in must

equal all work out

Work in = work to push the crateWork out = P.E.crate + Wf + K.E.crate

Wpush = F x s = 100 lbs x 10 ft = 1000 ft lbsP.E. = mgh = 2.0 slugs x 32 ft/s2 x (10 sin 300) ft = 320 ft lbs

Wf = Ff x s = FN s = (mg x cos 300) sWf = 0.10 x 2.0 slug x 32 ft/s2 x 0.866 x 10 ft = 55.4 ft lbs

K.E. = ½ mv2 = ½ (2) v2 = v2

Work out = P.E.crate + Wf + K.E.crate

1000= 320 + 55.4 + v2

v = (1000 – 320 – 55.4)1/2 = 25 ft/s

Work & Energy ProblemsA boy pulls a 30 lb cart 20 feet with a rope making a 600 angle with the

horizontal. The tension is the rope is 25 lbs. Disregard friction. How much work is done ?

Tension (T)

Verticalcomponent

Horizontalcomponent

Recall that only the component of the applied force in the direction of the

motion does work.Work = F cos x s

W = 25 lbs x cos 600 x 20 = 43.3 ft lbs

How much work is done if = 0.2 andthe cart moves at constant speed?

Recall: Wf = Ff x s = FN s The normal force of the cart

is reduced by the upward pullof the rope.

Upward pull of rope = T sin 600

Pup = 25 lbs x 0.5 = 12.5 lbsFN = w – Pup = 30 – 12.5 = 17.5 lbs

Wf = Ff x s = FN sWf = 0.2 x 17.5 lbs x 20 ft = 70 ft lbs

Work & Energy ProblemsElectrons in a TV tube have a mass of 9.11 x 10-28 grams

and a velocity of 3 x 107 m/s. What is there kinetic energy?

e3 x 107m/s

K.E. = ½ mv2

Using MKS units (meters, kilograms, seconds), 9.11 x 10-28 grams = 9.11 x 10-31 kg.

K.E. = ½ (9.11 x 10-31 kg)(3 x 107 m/s)2

K.E. = 4.1 x 10-16 joules / electron

Work & Energy ProblemsA bug crawls up a flight of stair 2.0 meters high in 5.0

minutes. His mass is 7.0 grams. What is his power output?

2.0 m

P = W / tThe bug is lifting himself

P.E. = mghUsing the CGS system2.0 meters = 200 cm

5.0 minutes = 300 secondsW = P.E. = 7.0 g x 980 cm/s2 x 200 cm

W = 1.37 x 106 ergs1 joule = 10,000,000 ergs (107)

W = 0.137 joulesP = 0.137 j / 300 sec = 4.6 x 10-4 watts

Work & Energy ProblemsA spring is stretched 5.0 cm when a mass of 100 grams is hung on it. How much work is needed to spring the same spring from 2.0 cm to 7.0 cm ?

100grams

5.0 cm

2.0 cm

7.0 cm

Hooke’s LawF = - kx

k = F/x, F = mgk = (0.1 kg x 9.8 m/s2) / 0.05 m

k = 19.6 N/m

Work = Fdx W = ½ k (x2

2 – x12)

W = ½ (19.6 N/m)(0.0702 – 0.0202 )0.0441 joules

Work & Energy ProblemsThe spring in the previous problem is compressed 3.0 cm lying on the horizontal. It is released against a 50.0 gram toy cart. What is the speed of the cart leaving the spring?

v

Work stored in spring is released as kinetic energy to the cartWspring = K.E. of cart

0.0441 joules = ½ mv2

50.0 grams = 0.050 kg (MKS)0.0441 joules = ½ (0.050 kg) v2

v = (0.0441 / 0.025)1/2 = 1.33 m/s or 133 cm/sec

Work & Energy ProblemsA 3000 lb car ascends a 15 0 hill at a constant speed of 30

mph. What is the power output of the car ? (Neglect friction)

30 mph = (30 x 5280) / 3600 = 44 ft /s In 1.0 second, distance = 44 ft

Work in = Work out Work in = lifting car + K.E. of car

Wf = 0P.E. = mgh ,w = mg

h = d x sin 15 0

P.E. = 3000 lbs x 44 ft x sin 15 0

P.E. =34,200 ft lbsK.E. = ½ mv2

At constant speed, K.E.=0

Work in = lifting car + K.E. of car W = 34,200 + 0 = 34,200 ft lbsP = W / t = 34,200 ft lbs / 1 sec

1 hp = 550 ft lbs / sec34,200 / 550 = 62.2 hp

150

h

30 mph 3000 lbs

d

Find the power output of an 80 lb girl who climbs a 12 ft ropein 7.5 seconds ?

(A) 0.13 hp (B) 0.23 hp (C) 2.0 hp (D) 960 hp

What is the velocity of a car with kinetic energy of 360 kj ? The mass ofThe car is 5.0 metric tons.

(A) 11.9 m/s (B) 36 m/s (C) 1.2 m/s (D) 24 m/s

A mass of 16 slug is elevated 20 feet. What is its change in potential energy?(A) 510 watts (B) 320 joules (C) 320 ft lbs (D) 5120 ft lbs

A spring is stretched from its normal length by 30 mm using a force of 0.40 N. How energy is stored in the spring?

(A) 6.0 x 10-3 j (B) 3.0 x 10-5 j (C) 7.2 x 10-2 j (D) none of these

A man pushes with a force of 200 N and moves a box 8 meters up a 200

incline . How much work does he do ?(A) 1600 j (B) 1500 j (C) 550 j (D) 200 ft lbs

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