Work and Energy
Energy

KINETIC ENERGY (KE): energy
associated with motion of an object

POTENTIAL ENERGY (PE): energy
associated with position of an object
WORK

WORK: the product of the component of the
force along the direction of displacement and
the magnitude of the displacement.
• W=F • dcosq
where d = magnitude of displacement
• When the force is not in the direction of the
displacement, then either sin or cos is needed.
• Units of work: 1 Joule = 1 N•m
In order to do work, you must exert
a force over a distance.
The con may
expend energy when
he pushes on the
wall, but if it doesn’t
move, no work is
performed on the
wall.
The force must be in the same direction as the
movement. If not, you have to use a vector
component of the force that is in the same
direction. Hence the cosine.
A lot of the time, you will
see the work equation
written like this:
W = FD cos q
Why?
Example: A man cleaning his apartment pulls
a vacuum cleaner with a force of 50. N at an
angle of 30.0. A frictional force of
magnitude 40. N dampens the motion, and the
vacuum is pulled a distance of 3.0 m.
Calculate:
a)the work done by the 50. N pull;
b)the work done by the frictional force;
c)the net work done on the vacuum by all
forces acting on it.
Example: A man cleaning his apartment pulls a vacuum cleaner with a force
of 50. N at an angle of 30.0. A frictional force of magnitude 40. N
dampens the motion, and the vacuum is pulled a distance of 3.0 m.
Calculate:
a)the work done by the 50 N pull;
b)the work done by the frictional force;
c)the net work done on the vacuum by all forces acting on it.
WF = (Fcosq) • d
50 N
WF = (50. N)(cos 30)(3.0 m)
30
WF = 130 J
40 N
direction of displacement
Example: A man cleaning his apartment pulls a vacuum cleaner with a force
of 50. N at an angle of 30.0. A frictional force of magnitude 40. N
dampens the motion, and the vacuum is pulled a distance of 3.0 m.
Calculate:
a)the work done by the 50. N pull;
b)the work done by the frictional force;
c)the net work done on the vacuum by all forces acting on it.
Wf = (Fcosq) • d
50 N
Wf = (40. N)(cos 180)(3.0 m)
30
Wf = -120 J
40 N
direction of displacement
Example: A man cleaning his apartment pulls a vacuum cleaner with a force
of 50 N at an angle of 30. A frictional force of magnitude 40 N dampens
the motion, and the vacuum is pulled a distance of 3 m. Calculate
a)the work done by the 50 N pull;
b)the work done by the frictional force;
c)the net work done on the vacuum by all forces
*normal force,
acting on it.
Wnet = WF + Wf
Wnet = 130 J – 120
Wnet = 10 J
50 N
weight, & upward
component of pull do no
work because they are
J perpendicular to the
displacement.
30
40 N
direction of displacement
Work and Kinetic Energy
Wnet  F  d  mad
remember : v2  vi 2  2ad
v 2 - vi 2
or... ad 
2
 v 2 - vi 2 

so, w net  m
 2 


or,
Wnet
1
2 1
 mvf  mvi 2
2
2
Work and Kinetic Energy
Wnet
1
2 1
 mvf  mvi 2
2
2
Roughly speaking,
energy
1 2 is the ability to
Kinetic Energy  KE  mv
do work.
2
so,
Wnet  KE f - KE i
or,
Wnet  KE
This is known as the work-energy theorem.
Work-Energy Theorem

The net work done on an object by the
force or forces acting on it is equal to
the change in the kinetic energy of the
object.
– The speed of an object will increase if the
net work done on it is positive;
– The speed of an object will decrease if the
net work done on it is negative.
KEi + PEi + Pei(s) + Workadded = KEf + PEf + Pef(s)
+ Worklost
This is where we
are going with this
unit.
We will now break
this equation down
piece by piece.
Note: Energy can be in all four or any combination of the
four types on either side. You need to decide which
pieces you will use.
Example: A 1400-kg car has a net forward force of 4500 N
applied to it. The car starts from rest and travels down a
horizontal highway. What are its kinetic energy and speed
after it has traveled 100.m? (ignore friction and air resistance)
m = 1400 kg
Wnet  KE
F = 4500 N
F  d  KE f  KEi
Vo = 0
d = 100. m
KEf = ?
(4500 N )(100.m)  KE f  0
KE f  450,000  4.5 10 J
5
KEi = 0
Vf = ?
4500 N
100 m
Example: A 1400-kg car has a net forward force of 4500 N
applied to it. The car starts from rest and travels down a
horizontal highway. What are its kinetic energy and speed
after it has traveled 100.0m? (ignore friction and air
resistance)
KEf  mvf
1
2
2
4.5 10  (1400kg)vf
5
1
2
vf  25 m/s
4500 N
100 m
2
Potential Energy

Gravitational potential energy: the
energy an object has due to its position
in space.
PE = mgh

In working problems, choose the zero level for
PE (the pt. at which the grav. PE=0) so that
you can easily calculate the difference in PE
Example: A 60.0-kg skier is at the top of a slope. At
the initial point A, the skier is 10.0 m vertically
above point B. Find the gravitational PE of the skier
at A and B, and the difference in PE between the 2
pts.
choose B = 0
PEi = mghi
PEi = (60.0kg)(9.8m/s2)(10.0m)
PEi =5880 J
PEf = mghf
PEf = (60.0kg)(9.8m/s2)(0m) B
PEf =0 J
A
Example: A 60.0-kg skier is at the top of a slope. At
the initial point A, the skier is 10.0 m vertically
above point B. Find the gravitational PE of the skier
at A and B, and the difference in PE between the 2
pts.
PEi =5880 J
PEf =0 J
A
PE = PEf - PEi
PE = -5880 J
B
Law of Conservation of Energy:
Energy can not be created or destroyed.
Actually, It’s
conservation of
mass - energy
Remember E=mc2
That is to say, energy
and mass are
interchangeable.
Although energy cannot be destroyed, it
can go places where we can never recover it.
Potential
Kinetic
Chemical
Energy
Heat
RIP
Electrical
Sound
And many more…
Conservation of Mechanical Energy
Although energy may be changed into a
different form, the final value will be the
same as the initial value (energy is not
lost).
Ei = Ef
KEi + PEi = KEf + PEf
½ mvi2 + mghi = ½ mvf2 + mghf
Kinetic vs. Potential Energy
Calculate the KE and PE energy at each of the 5 spots on the
ski run. Assume the skier starts from rest. (m = 51 kg)
Kinetic vs. Potential Energy
Calculate the KE and PE energy at each of the 5 spots on
the roller coaster. Assume the skater starts from rest.
Example: A diver weighing 750. N (mass =
77.0 kg) drops from a board 10.0 m above
the surface of the pool of water.
A) Use the conservation of
mechanical energy to find her
speed at a point 5.00 m above the
water surface.
10 m
B) Find the speed of the diver just
before she strikes the water.
Example: A diver weighing 750. N (mass = 77.0 kg) drops
from a board 10.0 m above the surface of the pool of water.
A) Use the conservation of mechanical energy to find
her speed at a point 5 m above the water surface.
B) Find the speed of the diver just before she strikes the water.
hf = 5 m
½ mvi2 + mghi = ½ mvf2 + mghf
½(77)(0)+(77)(9.8)(10)=½(77)(vf2)+(77)(9.8)(5)
Vf = 9.90 m/s
Example: A diver weighing 750 N (mass = 77.0 kg) drops from
a board 10 m above the surface of the pool of water.
A) Use the conservation of mechanical energy to find her speed
at a point 5 m above the water surface.
B) Find the speed of the diver just before she strikes
the water.
hf=0
½ mvi2 + mghi = ½ mvf2 + mghf
½(77)(0)+(77)(9.8)(10)=½(77)(vf2)+(77)(9.8)(0)
Vf = 13.9 m/s
Example: A sled and its rider together weigh 800. N.
They move down a frictionless hill through a vertical
distance of 10.0 m. Use conservation of energy to
find the speed of the sled-rider system at the bottom
of the hill, assuming the rider pushes off with an
initial speed of 5.00 m/s.
½ mvi2 + mgyi = ½ mvf2 + mgyf
½ vi2 + gyi = ½ vf2 + gyf
½(5)2+ (9.8)(10)=½(vf2)+ (9.8)(0)
vf = 14.9 m/s
10 m
Conservative / Nonconservative
Forces

A force is CONSERVATIVE if the work
it does on an object between 2 points is
independent of the path the object takes
between the 2 points. (The work done
depends only on the initial and final
positions.) (example: Work done by
gravity)
Conservative / Nonconservative
Forces

A force is NONCONSERVATIVE if the
work it does on an object moving
between 2 points depends on the path
taken. (i.e. Work done by sliding
friction)
Conservation of Energy with
Nonconservative Forces.
A 15 kg kid, initially at rest, slides down
a 8.0 m high slide. Ideally, what is his
final velocity at the bottom of the slide?
 KEi + PEi = KEf + PEf
 0 + mgh = 1/2 mv2 + 0
 (15kg)(9.8m/s2)(8.0m) = 1/2 (15kg)v2
 vf = 12.5m/s ~13m/s

In reality, the final velocity of the
kid is only 10 m/s. Where did
the extra energy go?
Friction is an nonconservative force which used
up some of the initial potential energy. To
account for friction, we need to add back in
the work done by (or energy lost by) friction.
KEi + PEi + Workadded = KEf + PEf + Worklost
Example:
Work of someone
pushing
Example:
Work done by
friction
Note: Usually work done by friction is negative (because
it is in the opposite direction). But since we are
ADDING BACK IN the work done by friction it is
positive.
How much work was done by
friction as the kid slid down the
slide? (m = 15 kg, yi = 8.0m vf = 10 m/s)
KEi + PEi + Workadded = KEf + PEf + Worklost
0 + mgh + 0 = 1/2mvf2 + 0 + Wfriction
(15kg)(9.8m/s2)(8.0m) = 1/2(15kg)(10.m/s)2 + Wf
1176 J = 750 J +Wf
Wf = 426 J ~ 430J
Potential Energy Stored in a Spring
PEs = ½ kx2
where k = spring constant
x = distance spring is compressed or stretched
(KE + PEg + PEs)i = (KE + PEg + PEs)f
Example: A block of mass 0.500 kg rests on a horizontal, frictionless surface.
The block is pressed lightly against a spring, having a spring constant k=80.0
N/m. The spring is compressed a distance of 2.00 cm and released. Find the
speed of the block at the instant it loses contact with the spring at the x = 0
position.
0.500 kg
(KE + PEg + PEs)i = (KE + PEg + PEs)f
(PEs)i = (KE)f
½kxi2 = ½mvf2
½(80)(0.02)2 = ½(0.5) vf2
v = 0.253 m/s
Example: The launching mechanism of a toy gun
consists of a spring of unknown spring constant. By
compressing the spring a distance of 0.120 m, the gun
is able to launch a 20.0-g projectile to a maximum
height of 20.0 m when fired vertically from rest.
Determine the value of the spring constant.
(KE + PEg + PEs)i = (KE + PEg + PEs)f
(PEs)i = (PEg)f
½kxi2 = mgh
½(k)(0.12)2 = (0.02)(9.8)(20)
k = 544 N/m
Machines make work easier
Work In = Work Out + Heat
Work Out
Efficiency 
 100
Work In
In a real machine the efficiency
must always be less than 100%
Types of Simple Machines:
Lever
 Pulley
 Ramp

Work In
Fulcrum
Distance Out
Distance In
Work In = Work Out
Force x
Distance = Force x
Distance
Work Out
25 lbs
25 lbs
50 lbs
50 lbs
50 lbs
Machines make work easier.
50 lbs
25 lbs
He lifts the
package up 4
ft…
4 ft
4 ft
4 ft
8 ft
But the work stays the same.
Remember…W = F x d
Power

Power: the rate at which work is done
W Fd


 F v
P
t
t
A bulldozer and a small boy can do
the same amount of work. It’s just
that the bulldozer can do it faster.
Units in the SI system….
Work Joules
Power 
 Watts

sec
time
Units in the English system
ft  lbs
Power 
sec
550 ft  lbs
 1 horsepower
sec
1 HP = 746 Watts
A new conveyor system at the local packaging plan will utilize a motorpowered mechanical arm to exert an average force of 890 N to push large
crates a distance of 12 meters in 22 seconds. Determine the power output
required of such a motor.

490 W (rounded from 485 W)
The Taipei 101 in Taiwan is a 1667-foot tall, 101-story skyscraper. The
skyscraper is the home of the world’s fastest elevator. The elevators
transport visitors from the ground floor to the Observation Deck on the
89th floor at speeds up to 16.8 m/s. Determine the power delivered by the
motor to lift the 10 passengers at this speed. The combined mass of the
passengers and cabin is 1250 kg.

2.06 x 105 W
Example: A 2.00x103 kg car starts from rest and
accelerates to a final velocity of +20.0 m/s in a time
of 15.0 s. Assume that the force of air resistance
remains constant at a value of –500. N during this
time
A)
B)
Find the average power developed by
the engine.
Find the instantaneous power when
the car reaches its final velocity
Example: A 2000 kg car starts from rest and accelerates to a final velocity
of +20.0 m/s in a time of 15.0 s. The force of the engine is 3160 N.
A) Find the average power developed by the engine
(average power uses average v).

 

vf  vi  at
vi  0 m/s

20.0 m/s  (a)(15.0 s)
vf  20.0 m/s
t  15.0 s
 F  Fengine  Fair resistance
a  1.33m/s2

 ma
Feng  500. N  (2000. kg)(1.33m/s )
2
Feng  3160 N
Example: A 2000 kg car starts from rest and accelerates to a final velocity
of +20 m/s in a time of 15 s. Assume that the force of air resistance
remains constant at a value of –500 N during this time.
A) Find the average power developed by the engine
(average power uses average v). Feng  3160 N
v f  vi

vavg 
2
20.0 m/s  0 m/s

vavg 
2

vavg  10.0 m/s

4
3
.
16

10
W

(
3160
N)(10.0
m/s)

P  F  vavg
1 hp
P  3.16  10 W 
 42.4 hp
746 W
4
Example: A 2000 kg car starts from rest and accelerates to a final velocity
of +20 m/s in a time of 15 s. Assume that the force of air resistance
remains constant at a value of –500 N during this time.
B) Find the instantaneous power when the car
reaches its final velocity (use velocity at instant you
are solving for).
P  F  vf
P  (3160 N)(20.0 m/s)
P  6.32  104 W
1 hp
P  6.32  10 W 
 84.8 hp
746 W
4
Example: An elevator has a mass of 1000. kg and carries a
maximum load of 800. kg. A constant frictional force of
4000. N retards its motion upward. What minimum power
must the motor deliver to lift the fully-loaded elevator at a
constant speed of 3.00 m/s? (In watts and horsepower)
v is constant, so a=0
FT – Ff – mg = 0
FT = Ff + mg = 4000 N + (1800 kg)(9.8m/s2)
FT = 2.16 x 104 N
P = FT v
P = (2.16 x 104 N)(3.00 m/s) = 6.48 x 104 W
P = (6.48 x 104 W)(1 hp / 746 W) = 86.9 hp