Transcript work
TODAY’S LESSON Learn what “work” is!
Learn how to calculate work See who can do the most work!
Learn about power.
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
a force does
work
when on it.
WORK In physics, English.
work has a special meaning, different to “normal”
BUT WHAT IS WORK?
A force does (and changes its energy) when it causes a
work
on a body
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
displacement, such as the normal force, can
do no work
.
to 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
m a
If
m
= 5 kg
a
= 10 m·s -1
F
= 50 N
DOT PRODUCT
W
A
A
is the
B
Note that the dot product of two vectors gives a scalar .
B
angle
AB
between cos
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 F
θ 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?
F
2 θ
F
1
VECTORS AND WORK
F
VECTORS AND WORK
F s
W =
F • s
W = F s cos 0 o W = F s 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 180 o W = - F s 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
m
g
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 frictional force is
F N
, the
f k
, and the weight of the box is
mg
.
Decide which forces do positive, zero, or negative work.
POSITIVE, ZERO, OR NEGATIVE WORK?
v
F N f k P mg
s
UNITS OF WORK
Energy is measured in Joules (J).
That’s me!
J = N·m J = kg·m
2
·s
-2
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) W = x a x b F(x) dx x a x b 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?
CAN YOU COPY THIS PLEASE?
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) Power Work Done time
P
W 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
125
kx
N
k
2 .
8 cm
k
125 N 2 .
8 cm 44 .
6 N cm Step 2: Solve for the force
T T T F
kx
44 .
6 N 98 N cm 98 N 3 .
2 cm
ELASTIC LIMIT