learn what “work” is! learn how to calculate work see who can do the most work! learn about...
TRANSCRIPT
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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.
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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.
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WORK
In physics, work has a special meaning, different to “normal” English.
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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.
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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?
Nad
a ZipZilch
NONEZERO
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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.
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CALCULATING WORK Work is the dot product of force and displacement.
Work is a scalar resulting from the interaction of two vectors.
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VECTOR MULTIPLICATION
There are three ways to multiply vectors:
•Scalar Multiplication
•Dot Product
•Cross Product
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SCALAR MULTIPLICATION
•Magnitude of vector changes.•Direction of vector does not change.
amF
a = 10 m·s-1
F = 50 N
If m = 5 kg
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DOT PRODUCT
BAW
Note that the dot
product of two vectors gives a scalar .
. and between angle theis BA
cosABBA
θ
A
B
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DOT PRODUCT
Geometrically, the dot product is the projection of one vector on a second
vector multiplied by the magnitude of the second vector.
θ
A
B
cosA
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CALCULATING WORK
cosFssFW
dxxFW )(
θ
F
s
cosF
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WHICH DOES MORE WORK?
θ
F1
F2
Two forces are acting on the box shown causing it to move across the floor. Which force does more work?
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VECTORS AND WORK
F
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VECTORS AND WORK
Fs
W = F • sW = F s cos 0o
W = F sMaximum positive work
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VECTORS AND WORK
F
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VECTORS AND WORK
s
W = F • sW = F s cos Only the component of force aligned with displacement does work. Work is less.
F
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VECTORS AND WORK
F
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VECTORS AND WORK
F s
W = F • sW = F s cos 180o
W = - F sMaximum negative work.
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GRAVITY OFTEN DOES NEGATIVE WORK.
mg
F
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.
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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.
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POSITIVE, ZERO, OR NEGATIVE WORK?
v
mg
P
FN
fk
s
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UNITS OF WORK
J = N·m
J = kg·m2·s-2
That’s me!
Energy is measured in Joules (J).
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WORK AND VARIABLE FORCEThe area under the curve of a graph of force vs displacement gives the work done by the force.
F(x)
xxa xb
W = F(x) dxxa
xb
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Let’s look at some examples
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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?
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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 = 400x15x0.985W = 5900 J
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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?
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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
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HOW MUCH WORK CAN YOU DO?
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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)
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ARM CURLS
distance
Force required = weight of object = mass (kg) x 10
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OFF YOU GO!
Name Mass (kg)
Force (N)
Distance (m)
Work of one lift (J)
# of lifts in 1 min
Total work (J)
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POWER!
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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)
t
WP
time
DoneWork Power
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POWER
For each of the people in your table, can you calculate their power?
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HOOKE’S LAW
When we stretch or compress a spring, a force arises that attempts to return the spring to its original length.
kxT
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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
kxT cm 8.2N 125 k
cmN6.44
cm 8.2
N 125k
Step 2: Solve for the force
kxT cm 2.36.44 cm
N TN98TN98F
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ELASTIC LIMIT