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Work and Energy
An Introduction
Work



Work tells us how much a force or
combination of forces changes the energy of
a system.
Work is the bridge between force (a vector)
and energy (a scalar).
W = F d cos 



F: force (N)
d : displacement (m)
: angle between force and displacement
Units of Work


SI System: Joule (N m)
 1 Joule of work is done when 1 N acts on a
body moving it a distance of 1 meter
British System: foot-pound


cgs System: erg (dyne-cm)



(not used in this class)
(not used in this class)
Atomic Level: electron-Volt (eV)
Electrical Work: kilo-Watt-hour (kW-hr)
Force and direction of motion
both matter in defining work!


There is no work done by a force if it causes
no displacement.
Forces can do positive, negative, or zero
work. When an box is pushed on a flat floor,
for example…



The normal force and gravity do no work, since
they are perpendicular to the direction of motion.
The person pushing the box does positive work,
since she is pushing in the direction of motion.
Friction does negative work, since it points
opposite the direction of motion.
Conceptual Checkpoint
Question: If a man holds a 50 kg box at
arms length for 2 hours as he stands
still, how much work does he do on the
box?
 Answer: NONE AT ALL

Conceptual Checkpoint
Question: If a man holds a 50 kg box at
arms length for 2 hours as he walks 1
km forward, how much work does he do
on the box?
 Answer: NONE AT ALL

Conceptual Checkpoint
Question: If a man lifts a 50 kg box 2.0
meters, how much work does he do on
the box?
 Answer:
W = Fd cos 
= (50 kg)(10 m/s2)(2.0 m)(cos 0o)
= 1,000 kg m2/s2
= 1,000 J

Work and Energy
Work changes mechanical energy!
 If an applied force does positive work
on a system, it increases mechanical
energy.
 If an applied force does negative work,
it decreases mechanical energy.
 The two forms of mechanical energy
are called potential and kinetic energy.

Sample problem #1
Jane uses a vine wrapped around a pulley
to lift a 70-kg Tarzan to a tree house 9.0
meters above the ground.
a) How
much work does Jane do when she
lifts Tarzan?
b) How much work does gravity do when Jane
lifts Tarzan?
Solution
a)
WJ = F d cos = (mg)Dx cos0o
WJ = (70 kg)(9.8 m/s2)(9.0 m)(1)
WJ = 6174 kg m2/s2 = 6174 J
(Jane does positive work because the force she
exerts is upward when it gets to Tarzan, and his
displacement is also upward.)
b)
WG = -6174 J
(Gravity pulls down but Tarzan goes up, so the
work done by gravity is negative.)
Sample problem
Joe pushes a 10-kg box and slides it
across the floor at constant velocity of
3.0 m/s. The coefficient of kinetic
friction between the box and floor is
0.50.
a) How
much work does Joe do if he pushes
the box for 15 meters?
b) How much work does friction do as Joe
pushes the box?
Solution
a)
WJ = F d cos0o
= (kFN)d = (kmg)Dx
(The force Joe exerts must be equal in magnitude
to the friction for there to be constant velocity.)
b)
WJ = (0.5)(10 kg)(9.8 m/s2)(15.0 m)
WJ = 735 J
Wf = -735 J
(The work done by friction is opposite to the work
done by Joe.)
Sample problem #3
A father pulls his child in a little red wagon
with constant speed. The father pulls with a
force of 16 N for 10.0 m, and the handle of the
wagon is inclined at an angle of 60o above the
horizontal.
a) How much work does the father do on the
wagon?
b) How much work does friction do on the
wagon?
Solution
a) WJ = Ffatherd cos 
= (16.0 N)(10.0 m) cos 60o
= 80 J
b) Wf = fd cos  = Ffatherd cos 180o
= (16.0 N)(10.0 m) cos 180o
= -80 J
Kinetic Energy
Energy due to motion
 KE = ½ m v2

K: Kinetic Energy
 m: mass in kg
 v: speed in m/s


Unit: Joules
Sample problem
A 10.0 g bullet has a speed of 1.2 km/s.
a) What
is the kinetic energy of the bullet?
b) What is the bullet’s kinetic energy if the
speed is halved?
c) What is the bullet’s kinetic energy if the
speed is doubled?
Solution
a)
b)
KE = ½ m v2
= ½ (0.010 kg) (1,200 m/s)2
= 7200 J
KE = ½ m ( ½ v) 2 = ¼ Ka (one-fourth as big)
= ¼ (7200 J) = 1800 J
c)
KE = ½ m ( 2 v) 2 = 4 Ka (four times bigger)
= 4 (7200 J) = 28800 J
The Work-Energy Theorem
 The
net work due to all forces equals
the change in the kinetic energy of a
system.
 Wnet = DKE
 Wnet:
work due to all forces acting on an
object
 DKE: change in kinetic energy (KEf – KEi)
Sample problem
An 80-g acorn falls from a tree and lands on the ground
10.0 m below with a speed of 11.0 m/s.
a) What would the speed of the acorn have been if
there had been no air resistance?
b) Did air resistance do positive, negative or zero
work on the acorn? Why?
c) How much work was done by air resistance?
d) What was the average force of air resistance?
Solution
a)
b)
c)
d)
v2 = vo2 – 2gDy
v = [-(2)(9.8 m/s2)(-10.0 m)] 1/2
v = 14 m/s
Air resistance did negative work on the acorn, since it was in
the direction opposite the displacement.
Wnet = DKE
W G + WD = ½ m v 2
WD = ½ m v2 – WG = ½ m v2 - FGDx
WD = ½ m v2 - mg Dx
WD = ½ (0.080kg) (11 m/s)2 - (0.080kg)(9.8m/s2)(10.0m)
WD = 4.84 J - 7.84 J = –3.00 J
WD = F Dx cos  = -F Dx
F = -WD /Dx = 3.00 J /10.0 m = 0.30 N
Constant force and work



The force shown is a
constant force.
W = FDr can be used
to calculate the work
done by this force
when it moves an
object from xa to xb.
The area under the
curve from xa to xb
can also be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Variable force and work



The force shown is a
variable force.
W = FDr CANNOT be
used to calculate the
work done by this
force!
The area under the
curve from xa to xb
can STILL be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Springs


When a spring is stretched or compressed from its
equilibrium position, it does negative work, since the
spring pulls opposite the direction of motion.
Ws = - ½ k x2





Ws: work done by spring (J)
k: force constant of spring (N/m)
x: displacement from equilibrium (m)
The force doing the stretching does positive work
equal to the magnitude of the work done by the
spring.
Wapp = - Ws = ½ k x2
Springs: stretching
0
F(N)
200
m
100
00
m
x
Fs
-100
-200
1
2
3
4
Fs
5
x (m)
Ws = negative area
= - ½ kx2
Fs = -kx (Hooke’s Law)
Sample problem
A spring with force constant 2.5 x 104 N/m
is initially at its equilibrium length.
a) How
much work must you do to stretch the
spring 0.050 m?
b) How much work must you do to compress it
0.050 m?
Solution
a)
b)
Wapp = ½ k x2
Wapp = ½ (2.5 x 104 N/m) (0.050 m)2
Wapp = 31.25 J
The same amount of work is needed to compress the
spring as to stretch it.
Sample problem
It takes 130 J of work to compress a
certain spring 0.10 m.
a) What
is the force constant of the spring?
b) To compress the spring an additional 0.10
m, does it take 130 J, more than 130 J, or
less than 130 J? Verify your answer with a
calculation.
Solution
a)
b)
Wapp = ½ k x2
130 J = ½ k (0.10 m)2
k = 2 (130 J) / (0.10 m)2
k = 26,000 N/m
To go from 0.10 m to 0.20 m, you need more than
130 J, since you need more and more force to
compress the spring the more you compress it.
Wapp = ½ k xf2 – 130 J
Wapp = ½ (26,000 N/m)(0.20m)2 – 130 J
Wapp = 520 J – 130 J = 390 J
Sample Problem

How much work is done by the force shown
when it acts on an object and pushes it from
x = 0.25 m to x = 0.75 m?
Solution
W = (0.40 N)(0.25 m) + (0.80 N)(0.25 m)
 W = 0.30 J

Sample Problem

How much work is done by the force
shown when it acts on an object and
pushes it from x = 2.0 m to x = 4.0 m?
Solution
W = ½ (base)(height)
 W = ½ (2 m)(0.60 N) = 0.60 J

Power
Power is the rate of which work is done.
 P = W/Dt

W: work in Joules
 Dt: elapsed time in seconds

When we run upstairs, t is small so P is
big.
 When we walk upstairs, t is large so P is
small.

Unit of Power

SI unit for Power is the Watt.
1 Watt = 1 Joule/s
 Named after the Scottish engineer James
Watt (1776-1819) who perfected the
steam engine.


British system
horsepower
 1 hp = 746 W

How We Buy Energy…
The kilowatt-hour is a commonly used
unit by the electrical power company.
 Power companies charge you by the
kilowatt-hour (kWh), but this not
power, it is really energy consumed.
 1 kW = 1000 W
 1 h = 3600 s
 1 kWh = 1000J/s • 3600s = 3.6 x 106J

Sample problem
A record was set for stair climbing when
a man ran up the 1600 steps of the
Empire State Building in 10 minutes and
59 seconds. If the height gain of each
step was 0.20 m, and the man’s mass was
70.0 kg, what was his average power
output during the climb? Give your
answer in both watts and horsepower.
Solution
P = W/t
= F Dx/t
= m g Dx/t
= (70.0 kg)(9.8 m/s2)[(0.20 m)(1600)] / (659 s)
= 333 W
and in horsepower
= 333 W / 746
= 0.45 hp
Sample problem
Calculate the power output of a 1.0 g fly
as it walks straight up a window pane at
2.5 cm/s.
Solution
P=W/t
Choose 1 second as the time frame.
At a velocity of 2.5 cm/s, the fly will go 0.025 m in 1 second.
P = FDx/t
= mg Dx/t
= (0.001 kg)(9.8 m/s2)(0.025) / 1 s
= 0.000245 W
Force types
Forces acting on a system can be
divided into two types according to how
they affect potential energy.
 Conservative forces can be related to
potential energy changes.
 Non-conservative forces cannot be
related to potential energy changes.
 So, how exactly do we distinguish
between these two types of forces?

Conservative forces

Work is path independent.



Work along a closed path is zero.



If the starting and ending points are the same, no
work is done by the force.
Work changes potential energy.
Examples:



Work can be calculated from the starting and ending
points only.
The actual path is ignored in calculations.
Gravity
Spring force
Conservation of mechanical energy holds!
Non-conservative forces

Work is path dependent.




Work along a closed path is NOT zero.
Work changes mechanical energy.
Examples:



Knowing the starting and ending points is not
sufficient to calculate the work.
Friction
Drag (air resistance)
Conservation of mechanical energy does not
hold!
Potential energy
Energy of position or configuration
 “Stored” energy
 For gravity: Ug = mgh

m: mass
 g: acceleration due to gravity
 h: height above some arbitrary “zero” point


For springs: Us = ½ k x2
k: spring force constant
 x: displacement from equilibrium position

Conservative forces and
Potential energy




Wc = -DU
If a conservative force does positive work on
a system, potential energy is lost.
If a conservative force does negative work,
potential energy is gained.
For gravity


Wg = -DUg = -(mghf – mghi)
For springs

Ws = -DUs = -(½ k xf2 – ½ k xi2)
More on paths and
conservative forces.
Q: Assume a conservative force
moves an object along the various
paths. Which two works are
equal?
A:
W2 = W3
(path independence)
Q: Which two works, when added
together, give a sum of zero?
A:
W1 + W2 = 0
or
W1 + W3 = 0
(work along a closed path is zero)
Sample problem
A box is moved in the
closed path shown.
a) How much work is
done by gravity
when the box is
moved along the
path A->B->C?
b) How much work is
done by gravity
when the box is
moved along the
path A->B->C->D->A?
Solution
WG = 0 + FDr
WG = 0 –mgh = -mgh
b)
WG = 0 -mgh + 0 + mgh
=0
The work in b) is zero
because work along a
closed path is zero for
any conservative
force.
a)
Sample problem
A box is moved in the
closed path shown.
a)
How much work
would be done by
friction if the box
were moved along
the path A->B->C?
b)
How much work is
done by friction
when the box is
moved along the
path A->B->C->D->A?
Solution
Wf = -kmgd - kmgd
Wf = -2kmgd
b) Wf = -kmgd - kmgd
- kmgd - kmgd
= -4 kmgd
Because friction is a
nonconservative
force, work along
the closed path in b)
is not zero.
a)
Sample problem
As an Acapulco cliff diver drops to the
water from a height of 40.0 m, his
gravitational potential energy decreases
by 25,000 J. How much does the diver
weigh?
Solution
DUG = mghf - mghi
DUG = mg(hf - hi)
mg = DUG / (hf - hi)
= -25,000 J /– 40.0 m
= 625 N
Sample problem
If 60.0 J of work are required to
stretch a spring from a 2.00 cm
elongation to a 5.00 cm elongation, how
much work is needed to stretch it from a
5.00 cm elongation to a 8.00 cm
elongation?
Solution
We need to find k for the spring first.
DUS = ½ kxf2 - ½ kxi2
DUS = ½ k(xf2 - xi2)
k = 2DUS / (xf2 - xi2)
= 2(60) / (0.052 – 0.0222) = 57,000 N/m
Now that we know k, we can calculate DUS.
DUS = ½ k(xf2 - xi2)
= ½ (57,000)(.082 – .052)
= 111 J
Law of Conservation of Energy
In any isolated system, the total
energy remains constant.
 Energy can neither be created nor
destroyed, but can only be transformed
from one type of energy to another.

Law of Conservation of
Mechanical Energy

E = KE + U = Constant
K: Kinetic Energy (1/2 mv2)
 U: Potential Energy (gravity or spring)


DE = DU + DKE = 0
DK: Change in kinetic energy
 DU: Change in gravitational or spring
potential energy

Skate Park Simulation
Skate Park Physics Simulation
(demonstrates gravitational potential energy,
kinetic energy, and thermal energy)
Pendulums and Energy
Conservation
Energy goes back and forth between KE
and Ug.
 At highest point, all energy is Ug.
 As it drops, Ug goes to KE.
 At the bottom , energy is all KE.

Pendulum Simulation at PHET website
Mechanical Energy Simulations
Energy Skate Park
Pendulum Lab
Masses and Springs
Pendulum Energy
½mvmax2 = mgh
h
For minimum and
maximum points of
swing
KE1 + U1 = KE2 + U2
For any points 1 and 2.
Springs and Energy
Conservation
Transforms energy back and forth
between KE and U.
 When fully stretched or extended, all
energy is U.
 When passing through equilibrium, all
its energy is KE.
 At other points in its cycle, the energy
is a mixture of U and KE.

Spring Energy
0
KE1 + U1 = KE2 + U2 = E
All U
m
For any two points 1 and 2
-x
All K
m
All U
m
x
½kxmax2 = ½mvmax2
For maximum and minimum
displacements from
equilibrium
Spring Simulation
Spring Physics Simulation
Sample problem
What is the
speed of the
pendulum bob
at point B if it
is released
from rest at
point A?
40o
1.5 m
A
B
1.5 cos 40o
40o
1.5 m
h
Ui + KEi = Uf + KEf
mgh + 0 = ½ mv2 + 0
v = 2gh
v = 2(9.8)[1.5(1-cos40o)]
v = 2.62 m/s
1.5 m
Solution
B
h = 1.5 – 1.5 cos 40o
h = 1.5 (1 - cos 40o)
A
Sample problem
A 0.21 kg apple falls from a tree to the
ground, 4.0 m below. Ignoring air
resistance, determine the apple’s
gravitational potential energy, U, kinetic
energy, K, and total mechanical energy,
E, when its height above the ground is
each of the following: 4.0 m, 2.0 m, and
0.0 m. Take ground level to be the point
of zero potential energy.
Solution – for 4.0 m
E = U + KE
= mgh + 0 (dropped from rest)
= (0.21 kg)(9.8 m/s2)(4.0 m) = 8.2 J
Therefore
E = 8.2 J (should be 8.2 J for entire problem)
U = 8.2 J(maximum value)
KE = 0 (minimum value)
Solution - for 2.0 m
E = U + KE
= mgh + Ke
8.2 J= (0.21 kg)(9.8 m/s2)(2.0 m) + KE
8.2 J = 4.1 J + KE
KE = 4.1 J
Therefore
E = 8.2 J
U = 4.1 J
KE = 4.1 J
Solution - for 0.0 m
E = U + KE
= mgh + KE
8.2 J = 0 + KE
8.2 J = KE
Therefore
E = 8.2 J
U=0
KE = 8.2 J
Sample problem
A 1.60 kg block slides with a speed of
0.950 m/s on a frictionless, horizontal
surface until it encounters a spring with a
force constant of 902 N/m. The block
comes to rest after compressing the
spring 4.00 cm. Find the spring potential
energy, U, the kinetic energy of the block,
KE, and the total mechanical energy of
the system, E, for the following
compressions: 0 cm, 2.00 cm, 4.00 cm.
Solution – for 0 cm
The spring has no energy, so all energy is
kinetic energy of the block.
E = KE
= ½ m v2 = ½ (1.60 kg)(.950 m/s)2
= 0.722 J
Therefore
E = 0.722 J(should be 0.722 J for entire problem)
U = 0 (minimum value)
KE = 0.722 J (maximum value)
Solution – for 4.00 cm
The block has stopped, so all energy is spring
potential energy!
Therefore
E = 0.722 J
U = 0.722 J
KE = 0
Solution – for 2.00 cm
The energy is a mixture of KE and U, but the
total energy is known. The spring potential
energy can be calculated.
E = U + KE = ½ k x2 + ½ m v2
0.722 J = ½ (902 N/m) (.0200 m)2 + KE
0.722 J = 0.180 J + KE
KE = 0.722 J – 0.180 J = 0.540 J
Therefore
E = 0.722 J
U = 0.1804 J
KE = 0.540 J
Law of Conservation of Energy

E = U + KE + Eint= Constant

Eint is thermal energy.
DU + DKE + D Eint = 0
 Mechanical energy may be converted to
and from heat.

Work done by nonconservative forces

Wnet = Wc + Wnc


Net work is done by conservative and non-conservative forces
Wc = -DU
• Potential energy is related to conservative forces only!

Wnet = DKE
• Kinetic energy is related to net force (work-energy theorem)

DKE = -DU + Wnc
• From substitution

Wnc = DU + DKE = DE

Nonconservative forces change mechanical energy. If
nonconservative work is negative, as it often is, the mechanical
energy of the system will drop.
Sample problem
Catching a wave, a 72-kg surfer
starts with a speed of 1.3 m/s, drops
through a height of 1.75 m, and ends
with a speed of 8.2 m/s. How much
non-conservative work was done on
the surfer?
Solution
Wnc = DU + DKE
= Uf – Ui + KEf – KEi
= mghf – mghi + ½ mvf2 – ½ m vi2
= m[g(hf –hi) + ½ (vf2 –vi2)]
= 72[(9.8)(0 - 1.75) + ½ (8.22 – 1.32)]
= 1125 J
Sample problem
A 1.75-kg rock is released from rest at the
surface of a pond 1.00 m deep. As the rock falls,
a constant upward force of 4.10 N is exerted on
it by water resistance. Calculate the
nonconservative work, Wnc, done by the water
resistance on the rock, the gravitational
potential energy of the system, U, the kinetic
energy of the rock, KE, and the total mechanical
energy of the system, E, for the following depths
below the water’s surface: d = 0.00 m, d = 0.500
m, d = 1.00 m. Let potential energy be zero at the
bottom of the pond.
Solution – for 0.00 m
Wnc = FDr = 0
E = U + KE
= mgh + 0 = mgh
= (1.75 kg)(9.8 m/s2)(1.00 m) = 17.15 J
Therefore
Wnc = 0 (the rock hasn’t moved yet)
E = 17.15 J(will be reduced by the drag force of water)
U = 17.15 J (maximum value)
KE = 0 (minimum value)
Solution – for 0.50 m
Wnc = FDx = (-4.10 N)(0.50 m) = -2.05 J = DE
E = 17.15 J – DE = 17.15 J - 2.05 J = 15.1 J
E = U + KE = mgh + KE
15.1 J = (1.75 kg)(9.8 m/s2)(0.50 m) + KE
15.1 J = 8.6 J + KE
KE = 15.1 – 8.6 J = 6.5 J
Therefore
Wnc = -2.05 J
E = 15.1 J(reduced by the drag force of water)
U = 8.6 J (determined by height)
KE = 6.5 J (reduced by the drag force of water)
Solution – for 1.00 m
Wnc = FDx = (-4.10 N)(1.00 m) = -4.10 J = DE
E = 17.15 J – DE = 17.15 J - 4.10 J = 13.05 J
E = U + KE = 0 + KE = KE
13.05 J = KE
Therefore
Wnc = -4.10 J
E = 13.05 J (reduced by the drag force of water)
U = 0 (lowest point in problem)
KE = 13.05 J (maximum value)