Work and Energy
Product of Vectors
Product of Vectors

We know, how to add and subtract vectors.

Vectors are not ordinary numbers, so ordinary multiplication
is not directly applicable to vectors.

We will define two kinds of product of vectors:

Scalar product: yields a result that is a scalar quantity.

Vector product: yields another vector.
Scalar Product

Scalar product of two vectors is a scalar quantity.

Scalar product can be negative, positive or zero.

It can be expressed in two ways:

In terms of magnitudes of A and B and the angle between
them:
 
 
A  B  AB cos   A B cos 


In terms of components of A and B:
 
A  B  Ax Bx  Ay By  Az Bz
 ranges from 0° to 180°
Scalar Product

Scalar product is:

Positive, when  is between 0° to 90°

Negative, when  is between 90° to 180°


Zero? The scalar product of two perpendicular vectors is always
zero.
Scalar product obeys the communicative law of multiplication: the
order of the two vectors does not matter.
   
A B  B  A

AB cos  BA cos
Unit vectors are  to each other. Scalar product of unit vectors:
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  (1)(1) cos 0  1
iˆ  ˆj  iˆ  kˆ  ˆj  kˆ  (1)(1) cos 90  0
 
A  B  Ax Bx  Ay By  Az Bz
Work
Work


Everyday life concept of work: any activity that requires muscular or
mental effort, for example

Push a stalled car off the road

Complete MasteringPhysics homework in time
Physics concept of work:


The total work done on a particle by all forces that act on it
equals the change in its kinetic energy (quantity that is related
to the particle’s speed  even non-constant speed and nonconstant forces; we will learn this later)
The work done by a constant force on an object that is
undergoing a straight line displacement is given by
 
W  F S
Work

The work done by a constant force on an object that is undergoing a
straight line displacement is given by
 
W  F S

Definition of work is based on observations. You do work by exerting the
force on an object while that object moves from one place to another
(undergoes displacement):

You do more work if the force is greater

You do more work if displacement is greater
Work
 
W  F S



SI unit of work is Joule (J)
“Jewel”
1 Joule = (1 Newton) (1 meter);
1 J = 1 N·m
British unit of work is foot-pound (ft·lb)

Unit of force is pound, unit of distance is foot

Conversion: 1 J = 0.7376 ft·lb, 1 ft·lb = 1.356 J
James Joule
1818 - 1889
Work

You push a stalled car through a displacement S with a constant force F
in the direction of motion:
W  F S

You push a stalled car through a displacement S with a constant force F
at angle  to the direction of motion:
W  F  S cos

Only component of force in direction of car’s displacement is important
Work and Kinetic Energy
Work and Kinetic Energy

From the definition of work we know that the total work done on an
object is related to its displacement (changes in position).
 
W  F S

Work is also related to changes in the speed of the object.
Work and Kinetic Energy

Example: Block sliding on a frictionless table

Forces acting on a block: its weight, normal
force, and the force F exerted by the hand.
Work and Kinetic Energy

Example: Block sliding on a frictionless table

A.
The net force on a block is
in the direction of its
motion. From N2L, this
means that the block
speeds up. W=F·S also
tells us that the total work
will be positive.
B.
Here only the component
F·cos contributes to Wtotal.
The block speeds up as well.
Forces acting on a block: its weight, normal
force, and the force F exerted by the hand.
Work and Kinetic Energy

Example: Block sliding on a frictionless table

C.
The net force here opposes
the displacement. From
N2L, this means that the
block slows down. W=F·S
also tells us that the total
work will be negative.
D.
Here the net force is zero,
so the speed of the block
stays the same, and Wtotal
is zero.
Forces acting on a block: its weight, normal
force, and the force F exerted by the hand.
Work and Kinetic Energy

When an object undergoes a displacement:



object will "speed up" if Wtotal > 0,
object will "slow down" if Wtotal < 0,
object will "maintain the same speed“ (constant) if Wtotal = 0.
Work and Kinetic Energy

Consider a particle with mass m moving along the x-axis under the
action of a constant net force with magnitude F directed along the
positive x-axis.



Particle acceleration is constant, and by N2L: F=m·ax.
Suppose, speed changes from v1 to v2 while particle undergoes a
displacement S=x2-x1 from point x1 to x2.
1-D constant-acceleration equation:
v22  v12  2  a x  S
v22  v12
F  max  m
2S
v22  v12
ax 
S  x2  x1
2S
1 2 1 2
F  S  mv2  mv1
2
2
Work and Kinetic Energy
1 2 1 2
F  S  mv2  mv1
2
2

The product F·S is the work done by the net force. Thus, it is equal to the
total work Wtotal done by all the forces acting on a particle.
1 2
K  mv
2

Definition of Kinetic Energy:

Like work, kinetic energy of a particle is a scalar quantity: it depends on
particle’s mass and speed, not its direction of motion.


Car has the same kinetic energy when going north at 10m/s as
when going east at 10m/s.
Kinetic energy can never be negative; it’s zero when particle is at rest.
Work - Energy Theorem
F S 
1 2 1 2
mv2  mv1
2
2
Ki 
1 2
mvi
2
F  S  K 2  K1

Work done by the net force on a particle equals the change in the
particle’s kinetic energy:

Work - Energy Theorem:

When an object moves:





Wtot  K 2  K1  K
object will "speed up" if Wtotal > 0, K2 > K1
object will "slow down" if Wtotal < 0, K2 < K1
object will "maintain the same speed“ (constant) if Wtotal =0,
K2=K1
Speeds and distances must be measured in inertial frame of reference!
Kinetic energy and work have the same units (Joules,
2 or N·m):
1J  1N  m  1(kg  sm2 )  m  1kg  ms 2
Work and Kinetic Energy
Problem Solving Strategy

IDENTIFY the relevant concepts: The work-energy theorem is extremely
useful in situations where you want to relate a body’s speed at one point in its
motion to its speed at a different point.


This approach is less useful for problems that involve time, such as finding the time
it takes a body to go from point 1 to point 2. The reason is that the work-energy
theorem doesn’t involve time at all. For problems that involve time, it’s usually best
to use the relationships among time, position, velocity, and acceleration we learned
in 1-D motion.
SET UP the problem using the following steps:
1. Choose the initial and final positions of the body, and draw a free-body
diagram showing all the forces that act on the body.
2. Choose a coordinate system. (If the motion is along a straight line, it’s
usually easiest to have both the initial and final positions lie along the xaxis.)
3. List the unknown and known quantities, and decide which unknowns are
your target variables. In some cases the target variable will be the
body’s initial or final speed; in other cases it will be the magnitude of
one of the forces acting on the body.
Work and Kinetic Energy
Problem Solving Strategy

EXECUTE: Calculate the work done by each force.


Add the amounts of work done by each force to find the total work.


Note that kinetic energy involves mass, not weight; if you are given the
body’s weight, you’ll need to find the mass.
Finally, use the relationship Wtot=K2-K1 to solve for the target variable.


Be careful with signs! Sometimes it may be easier to calculate the vector
sum of the forces (the net force), then find the work done by the net force.
Write expressions for the initial and final kinetic energies, K1 and K2.


Be sure to check the sign of the work for each force; it must be positive if
the force has a component in the direction of the displacement, negative if
it has a component opposite the displacement, and zero if the force and
displacement are perpendicular.
Remember that the right-hand side of this equation is the final kinetic
energy minus the initial kinetic energy, never the other way around.
EVALUATE your answer: Check whether your answer makes physical
sense. A key item to remember is that kinetic energy can never be
negative. If you come up with a negative value of K, you’ve made a
mistake!
Kinetic Energy



The example with the hammerhead gives insight into the physical
meaning of kinetic energy.
The hammerhead was dropped from rest, and its kinetic energy when it
hits the I-beam equals the total work done on it up to that point by the
net force.
To accelerate a particle with mass m from rest (zero kinetic energy) up
to a speed v, the total work done on it must equal the change in kinetic
energy from zero to K=0.5·m·v2:
Wtot  K  0  K


K  1 mv2
2
Kinetic energy of a particle is equal to the total work that was
done to accelerate it from rest to its present speed.
Or from its present speed to rest!


Catch the ball right  pull your hand back, increasing distance to stop
the ball: ball does the work on your hand equal to the ball’s initial
kinetic energy Wtot=FS=0.5mv2.
Pulling hand back, you maximize the distance over which this force acts
and thus minimize the force on your hand.
Composite Systems







Man standing on frictionless roller skates on a level
surface, pushes against the rigid wall, setting himself in
motion to the right.
Forces acting on him: his weight W, upward normal
forces n1 and n2 exerted by the ground on his skates,
and the horizontal force F exerted on him by the wall.
No vertical displacement, so forces W, n1 and n2 do
NO work. Force F exerted on him by the wall is
horizontal force that accelerates him to the right, but his
hands don’t move. So this force doesn’t do work as well.
Where does the man’s kinetic energy come from?
This system is not a single point (or particle)
Various parts of the body interact with each other (here:
his hands are still, but his torso moves)
Total kinetic energy of this composite system of body
parts can change, even no work is done by forces
applied outside the system.
Work and Energy with Varying
Forces
Work and Energy with Varying Forces

We know that work done by a constant force on an object that is
undergoing a straight line displacement is given by
 
W  F S

What happens when force exerted on an object is NOT constant
and the object moves in path which is NOT straight?

Example: spring, stretched

More you stretch it, the harder you have to pull: thus the
force is non-constant

Let’s consider straight-line motion with non-constant force

One complication at a time!

Fx change along the x-axis (force depends on position)
Work and Energy with Varying Forces




Particle moves from x1 to x2; Fx depends on coordinate x.
Let’s divide the total displacement by small segments xa, xb, xc…
Total work done during segment xa: ~ by the average force Fa in this
segment multiplied by the displacement xa.
All segments:
W  Fa xa  Fb xb  ...  Ff x f   Fmxm
Work and Energy with Varying Forces


If number of segments is very large, segment's width x is very small
In the limit, the sum is integral
of Fx from x1 to x2.
x2
W  lim  Fm xm    Fx dx
x 0

x1
On a graph of force as function of
position, the total work done by this
force is represented by the area
under the curve between the initial
and final positions.
Varying x-component of force,
straight-line displacement
Work and Energy with Varying Forces

Let’s check
from x1 to x2.:
x it: if Fx is constant
x
2
2
x1
x1
W   Fx dx  Fx  dx  Fx ( x2  x1 )  F  S
Hooke’s Law

To keep ideal spring stretched by amount x beyond its initial length,
we need to apply on the end the force which is proportional to x:
Fx  k  x
Force required to stretch a spring
K – spring (force) constant, [N/m]

Example: Force constant k
Floppy toy spring: k=1 N/m
Car suspension spring: k=105 N/m
Robert Hooke
1635 – 1703
... lean, bent and ugly man ...
Hooke’s Law




To stretch a spring, you must do work
Suppose one end of a spring is fixed, you apply force on another end
That end moves, so the force does work
Work done by the force when spring elongation goes from zero to X:
X
X
1
W   Fx dx   kxdx  kX 2
2
0
0



Total work is ~ to square of final
elongation
Graph, Area under the curve: W 
1
 X  kX
2
If spring was initially already stretched a
distance x1, the work to stretch it to a
greater elongation x2:
x2
x2
1
1
W   Fx dx   kxdx  kx22  kx12
2
2
x1
x1
What happens if you
compress the spring?
Hooke’s Law
What happens if you compress the spring?
Compression:
 Force Fx and displacement x are both negative

Force is in the same direction as displacement: work is positive

Example is following…
 Fx
x
x
Varying Forces: Work - Energy
Theorem

One can use the same approach: divide total displacement into segments



Apply Work-Energy Theorem for each segment: Wa=Ka=Faxa
Sum the changes to find Wtot
Another way:
ax 
dvx
dx
, vx 
dt
dt
ax 

x2
x2
x2
x1
x1
x1
Wtot   Fx dx   max dx   mvx

“Chain rule” for derivatives
dv x
dx
dx
(dvx/dx)dx is the change in velocity dvx during displacement dx. Thus:
x2
Wtot   mvx dv x 
x1

dv x dvx dx
dv

 vx x
dt
dx dt
dx
1 2 1 2
mv2  mv1
2
2
Work – Energy Theorem is the same: valid for varying forces as well !
Curved Path: Work - Energy Theorem


Force that varies in direction and magnitude
Displacement lies along a curved path: particle moves from P1 to P2

Divide curve between P1 and P2 into small vector displacements d.

Each d is a tangent to the path at its position.

F is the force at a point along the path,  is the angle between F and d.

Small element of work W done on particle during displacement d :
dW  F cos d  F d  F  d
Total work done on a particle then:
P2
P2
P2
P1
P1
P1
W   F cos d   F d  

F  d
(Work done on a curved path)
Power
Watt’s
Engine
“Power man”
Mr. Olympia
Power
Definition of work makes no reference to the passage of time

You lift weight 100N vertically at a distance 1m at constant velocity:

You do (100N)(1m)=100J of work whether it takes 1 sec, 1 hour, 1
year…

You want to know how quickly the work is done

Power is the time rate at which work is done.

Average power:

W
Pav 
t
Instantaneous power:
W dW
P  lim

t 0 t
dt
Power is a scalar.
Power


The SI unit of power is watt (W), 1 W = 1 Joule per 1 second.
In the British system of units power is in ft·lb/sec or in a larger unit
called horsepower (hp).


1 hp = 550 ft·lb/sec = 33,000 ft·lb/min = 746 W = ¾ of kilowatt (kW)
The kilowatt-hour (kWh) is a unit of electrical energy, not power

energy = power · time
James Watt

In mechanics, power is expressed
in terms of force and velocity
Pav 
F S
W
S

F
 F vav
t
t
t
W dW

F v
t 0 t
dt
P  lim
1736 – 1819
Watt's steam engines
…Started with nothing, died
as a very wealthy man…
P  F v
In terms of scalar product
James Watt

Watt discovered how he could make the steam engine more efficient.




In 1755 Watt had been granted a patent by Parliament that prevented anybody
else from making a steam-engine like the one he had developed.
For next 25 years, the “Boulton & Watt” company had a virtual monopoly over
the production of steam-engines. These machines were mainly sold to colliery
owners who used them to pump water from their mines, and textile factories.
Watt charged his customers a premium for using his steam engines.


Steam engine that cooled the used steam in a condenser separate from main cylinder.
To justify this he compared his machine to a horse. Watt calculated that a horse
exerted a pull of 180 lb., therefore, when he made a machine, he described its power
in relation to a horse, i.e. "a 20 horse-power engine".
Watt worked out how much each company saved
by using his machine rather than a team of
horses. The company then had to pay him one
third of this figure every year, for the next
twenty-five years. When James Watt died in 1819
he was a very wealthy man.
Watt’s Steam Engine

This Watt Rotation Engine is housed in the Henry Ford Museum in Dearborn, MI