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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)
Applied
force
Vertical
component

Horizontal
component
Distance moved
Vertical component = Applied force x sin 
Horizontal component = Applied force x cos 
The horizontal force component is in the direction of the motion
Work = force in direction of motion x distance moved

FORCE OF FRICTION = 0
WORK DONE = 0
FORCE OF FRICTION > 0
WORK 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
Who are they
kidding ??
Three
If this point was used as reference,
One, Two and Three would all have
negative potential energies.
I sure am !
Boy Two!
You’re at a
High potential
Two
One
Two is at a higher potential
energy than One but lower than
Three. Two’s potential energy is
negative relative to Three’s and
positive relative to One’s.
1/9
wt
16.7 lbs
scale
Three Radius of Earth = 12000 miles
¼ wt
37.5 lbs
scale
Two Radius of Earth = 8000 miles
Normal
wt
150 lbs
scale
Radius of Earth = 4000 miles
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 = 0
No 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.
Elongation of spring
\
F
O
R
C
E
(N)
200
grams
400
grams
600
Slope = spring constant
ELONGATION (M)
grams
Constant force
Work = force x distance
Force x distance equals
area under the graph
F
o
r
c
e
Constant force
(N)
Distance (M)
Distance moved
x
Work = area under a
force versus distance graph
Area under the graph
gives the work to stretch
the spring. Work needed
to stretch the spring to x2
is ½ F2 times x2
Work needed to stretch
the spring to x1
is ½ F1 times x1
F F1
O
R
C
F1
E
(N)
Work needed to stretch
the spring from x1 to x2 is
(½ F2 times x2) – ( ½ F1 times x1)
X1
X2
ELONGATION (M)
Since F = kx and
W = ½ Fx, W = ½ (kx) x or
W = ½ kx2 and work from
x1 to x2 is given by:
W = ½ k (x22 – x12)

BOX METHOD
WORK = AREA UNDER THE CURVE
W =  F  X (SUM OF THE BOXES)
F
O
R
C
E
AREA
MISSED
- INCREASING
AS THE
NUMBER
OF BOXES
THE
NUMBERTHE
BOXES
WILL
INCREASES,
ERROR
REDUCE
THIS ERROR!
DECREASES!
(N)
WIDTH OF EACH BOX =  X
X1
X2
DISPLACEMENT (M)
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 ANTIDERIVATIVE
INTEGRATION IS THE REVERSE PROCESS OF
FINDING THE DERIVATIVE. IT CAN ALSO BE USED
TO 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
 is the symbol
for integration
 
DIVIDE THE
EQUATION
BY THE N + 1
ADD A
CONSTANT
APPLYING THE INTEGRAL FORMULA
GIVEN THE
EQUATION
FORMAT TO
FIND THE
INTEGRAL
 

Integration can be used to find area under a curve between
two points. Also, if the original equation is a derivate, then
the equation from which the derivate came can be determined.
APPLYING THE INTEGRAL FORMULA
Find 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 are
called the limits.
each of the limits is
placed in the integrated
equation and the results
of each calculation are
subtracted (lower limit
from upper limit)
Finding the Equation for Work Stored in
Spring using Integration
Hooke’s Law F = kx
Work =  Fdx =  kx dx
Work = k x1+1/ (1+1) = k x2 / 2 + C
x2
2
W=½ kx |
x1
Note that the work equation is the same
as that found by area under the curve
methods used in previous slides
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.
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.
ground
PE = mg h
High potential energy
Low kinetic energy
h
KE = ½ mv2
Low potential energy
High kinetic energy
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 Problems
A 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 ?
420 N
500 N
5m
Work = F cos  x s
Since the weight of the box
is not the applied force it is
not related to the work done.
Horizontal means that  = 00, cos 00 = 1.0
W = 420 N x 1.0 x 5 m = 2100 joules
(Recall : 1 j = 1 N x 1 m)
or 2.1 kilojoules
Work & Energy Problems
A 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 ?
? N
Ff
60 kg
w = mg
w = 60 x 9.8 = 588 N
12 m
Constant speed means no
acceleration 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 N
Work = F cos  x s,  = 00
W = 176 N x (1.0) x 12 = 2112 j
Work & Energy Problems
How 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?
1.5’
7.0 ft
5’6’’
Lifting always involves
changing gravitational
potential energy.
P.E. = mgh
weight is m x g
and is expressed in pounds.
P.E. = mgh
P.E. = 100 lbs x 7 ft = 700 ft-lbs
Work & Energy Problems
A 20.0 kg crate is pulled 8.00 meters up a frictionless
incline with a 200 angle. How much work is done ?
8m
h
200
h = 8 x sin 200 = 2.74 m
P.E. = mgh
P.E. = 20 kg x 9.8 m/s2 x 2.74 m
P.E. = 536 j
The crate is actually being
lifted against gravity.
Although it is pulled 8 m
it is lifted only the vertical
distance h.
How much work is used
If 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 j
Work total = P.E. + Wf
Work total = 536 j + 80 j = 616 j
Work & Energy Problems
A box slides down a frictionless incline 12 feet long with
an angle of 300. What is its speed at the bottom?
12 ft
h
300
The energy in the box is being
converted from potential (it is
elevated) 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 ft
v = (2 x 32 ft/s2 x 6 ft)1/2 = 19.6 ft/s
Work & Energy Problems
A 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.
Conservation of energy tells
us that all work in must
equal all work out
10 ft
h
300
100 lbs
Work in = work to push the crate
Work out = P.E.crate + Wf + K.E.crate
Wpush = F x s = 100 lbs x 10 ft = 1000 ft lbs
P.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) s
Wf = 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 Problems
A 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 ?
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 and
the cart moves at constant speed?
Recall: Wf = Ff x s =  FN s
The normal force of the cart
is reduced by the upward pull
of the rope.
Tension (T)
Vertical
component

Horizontal
component
Upward pull of rope = T sin 600
Pup = 25 lbs x 0.5 = 12.5 lbs
FN = w – Pup = 30 – 12.5 = 17.5 lbs
Wf = Ff x s =  FN s
Wf = 0.2 x 17.5 lbs x 20 ft = 70 ft lbs
Work & Energy Problems
Electrons 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?
3 x 107m/s
e
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 Problems
A 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/t
The bug is lifting himself
P.E. = mgh
Using the CGS system
2.0 meters = 200 cm
5.0 minutes = 300 seconds
W = P.E. = 7.0 g x 980 cm/s2 x 200 cm
W = 1.37 x 106 ergs
1 joule = 10,000,000 ergs (107)
W = 0.137 joules
P = 0.137 j / 300 sec = 4.6 x 10-4 watts
Work & Energy Problems
A 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 ?
5.0 cm
100
grams
2.0 cm
Hooke’s Law
F = - kx
k = F/x, F = mg
k = (0.1 kg x 9.8 m/s2) / 0.05 m
k = 19.6 N/m

Work = Fdx
W = ½ k (x22 – x12)
7.0 cm
W = ½ (19.6 N/m)(0.0702 – 0.0202 )
0.0441 joules
Work & Energy Problems
The 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 cart
Wspring =  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 Problems
A 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
h
3000 lbs
d
150
P.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 lbs
K.E. = ½ mv2
At constant speed, K.E.=0
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 = 0
Work in = lifting car + K.E. of car
W = 34,200 + 0 = 34,200 ft lbs
P = W / t = 34,200 ft lbs / 1 sec
1 hp = 550 ft lbs / sec
34,200 / 550 = 62.2 hp
Find the power output of an 80 lb girl who climbs a 12 ft rope
in 7.5 seconds ?
(A) 0.13 hp (B) 0.23 hp (C) 2.0 hp (D) 960 hp
Click
here for
answers
What is the velocity of a car with kinetic energy of 360 kj ? The mass of
The 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