### work

```TODAY’S LESSON
Learn what “work” is!
 Learn how to calculate work
 See who can do the most work!
 Learn Hooke’s Law.

ENERGY AND WORK
Energy - the ability of a
body or system of bodies to
perform work.
A body is given energy when
a force does work on it.
WORK
In physics, work has a
special meaning,
different to “normal”
English.
BUT WHAT IS WORK?
A
force does work on a body
(and changes its energy) when it
causes a displacement.
 If
a force causes no
displacement, it does no work.
RIDDLE ME THIS
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?
ZERO
NONE
COUNTERINTUITIVE
RESULTS
 There is no work done by a force if it
causes no displacement.
 Forces perpendicular to
displacement, such as the normal
force, can do no work.
 Likewise, centripetal forces never
do work.
CALCULATING WORK
Work
is the dot product of force
and displacement.
Work
is a scalar resulting from
the interaction of two vectors.
VECTOR MULTIPLICATION
There are three ways to multiply
vectors:
•Scalar Multiplication
•Dot Product
•Cross Product
SCALAR MULTIPLICATION
•Magnitude of vector changes.
•Direction of vector does not change.


F  ma
If m = 5 kg
a = 10 m·s-1
F = 50 N
DOT PRODUCT
Note that the dot
 
W  A  B product of two vectors
gives a scalar .
 
A  B  AB cos



 is theangle between A and B.

A
θ

B
DOT PRODUCT
Geometrically, the dot product is the
projection of one vector on a second
vector multiplied by the magnitude of
the second vector.

A
θ
A cos 

B
CALCULATING WORK
 
W  F  s  Fs cos

F
θ
F cos 

s
W   F ( x ) dx
WHICH DOES MORE WORK?
Two forces are acting on the box
shown causing it to move across the
floor. Which force does more work?
F2
θ
F1
VECTORS AND WORK
F
VECTORS AND WORK
F
s
W=F•s
W = F s cos 0o
W=Fs
Maximum positive work
VECTORS AND WORK
F
VECTORS AND WORK
F

s
W=F•s
W = F s cos 
Only the component of force aligned
with displacement does work. Work is
less.
VECTORS AND WORK
F
VECTORS AND WORK
F

s
W=F•s
W = F s cos 180o
W=-Fs
Maximum negative work.
GRAVITY OFTEN DOES NEGATIVE WORK.
When the load goes up, gravity does negative
work and the crane does positive work.
When the load goes down, gravity does
positive work and the crane does negative
work.
F
mg
POSITIVE, ZERO, OR NEGATIVE WORK?
A box is being moved with a velocity v
by a force P (parallel to v) along a
level floor. The normal force is FN, the
frictional force is fk, and the weight of
the box is mg.
Decide which forces do positive, zero,
or negative work.
POSITIVE, ZERO, OR NEGATIVE WORK?
v
FN
fk
P
mg
s
UNITS OF WORK
Energy is measured in Joules (J).
That’s me!
J = N·m
J=
2
-2
kg·m ·s
WORK AND VARIABLE FORCE
The area under
the curve of a
graph of force vs
displacement
gives the work
done by the force.
F(x)
xb
W = x F(x) dx
a
xa
xb
x
Let’s look at
some examples
WORK DONE (J) = FORCE (N) X DISTANCE (M)
A woman pushes a car with a force of 400 N
at an angle of 10° to the horizontal for a
distance of 15m. How much work has she
done?
WORK DONE (J) = FORCE (N) X DISTANCE (M)
A woman pushes a car with a force of 400 N
at an angle of 10° to the horizontal for a
distance of 15m. How much work has she
done?
W = Fscosθ = 400x15x0.985
W = 5900 J
WORK DONE (J) = FORCE (N) X DISTANCE (M)
A man lifts a mass of 120 kg to a height of
2.5m. How much work did he do?
WORK DONE (J) = FORCE (N) X DISTANCE (M)
A man lifts a mass of 120 kg to a height of 2.5m.
How much work did he do?
Force = weight = 1200N
Work = F x d = 1200 x 2.5
Work = 3000 J
HOW MUCH WORK CAN YOU DO?
Name
Mass
(kg)
Force
(N)
Distance
(m)
Work of
one lift
(J)
# of lifts in
1 min
Total work (J)
ARM CURLS
Force required = weight of object = mass (kg) x 10
distance
OFF YOU GO!
Name
Mass
(kg)
Force
(N)
Distance
(m)
Work of
one lift
(J)
# of lifts in
1 min
Total work (J)
POWER!
POWER!
Power is the rate of doing work.
Power is the amount of work done per unit time.
Power is measured in Watts (1 Watt = 1 J/s)
W ork Done
P ower 
t ime
W
P
t
POWER
For each of the people in your table, can you
calculate their power?
HOOKE’S LAW
When we stretch
or compress a
spring, a force
arises that
attempts to return
the spring to its
original length.
T  kx
OUR FIRST HOOKE’S LAW PROBLEM
A force of 125 N is required to extend a spring by
2.8 cm. What force is required to stretch the
same spring by 3.2 cm?
Step 1: Solve for k
T  kx
125 N  k  2.8 cm
125 N
N
k 
 44.6 cm
2.8 cm
Step 2: Solve for the force
T  kx
N
T  44.6 cm
 3.2 cm
T  98 N
F  98 N
ELASTIC LIMIT
```