Download Honors Work, Energy

Work, Energy, Power
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What’s the difference?
• Force is the agent of change
• Energy is a measure of change
• Work is a way of transferring energy
from one system to another
What is work?
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• Work=
force*displac.
• W=Fd
• Only work if there is
motion- if you push
against a brick wall
and it doesn’t move,
you might be tired
but you have done
no work
• Unit=Joule (unit of
energy)
Work and Force at an angle
• Force and work must
be in same direction so
if force is at angle to
displacement, must
use component of
force parallel to
displacement
• So W=Fdcos where 
is angle between F and
d
Are they work?
• Teacher pushes wall and
becomes exhausted
• Book falls off table to
floor
• Waiter carries large tray
across restaurant at
constant v
• Starship Enterprise
accelerates through
space
• No- no displacement
• Yes- force=g and
displacement=fall
• No-why?
• Yes- force from
engines
So what’s with the waiter???????
What’s with the waiter?
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• There is a force (the waiter
pushes up on the tray) and
there is a displacement (the
tray is moved horizontally
across the room).
• Yet the force does not
cause the displacement. To
cause a displacement,
there must be a component
of force in the direction of
the displacement.
• There was a small force
when the waiter
accelerated to start moving
but no net force to keep it
moving
+ and - work
• If a force helps the
motion, W is +
– cos is +
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• If force opposes motion,
W is – cos is -
• If force is perpendicular
to motion, W=0
– cos is 0
Is work done by gravity + or -?
Is work sone by person + or -?
Work = 0
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• Work = 0 if:
– No force
– No displacement
– cos =0 or force is
perpendicular to
displacement
Example…
• A box slides down an inclined plane (angle =
37 degrees). The mass of the block is 35kg
and the length of the ramp is 8m.
– What is work done by gravity?
– BE CAREFUL- angle given is NOT the angle
between F and d!!!
Solution
•
•
•
•
Wgavity=(mgsin )d
=(35kg)(10N/kg)(sin37)(8m)
=1680J
Note work done by gravity is + b/c it is
helping with motion
• Also- be careful with the angle  - usually we
measure angle between F and D but in this
case the angle given was the incline angle
Kinetic Energy and
Work-Energy Theorem
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• When you do net
work on an object,
the result is a
change in kinetic
energy
• K=1/2mv2
• Wtotal=K
Example…
• A tennis ball of mass
0.06kg is hit straight up
at an initial speed of
50m/s. How high would
it go?
• You can solve this using
the big 5 or using the
work-energy theorem
using gravity as the
force doing negative
work
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Solution using WORK
•
•
•
•
•
•
•
•
Force= gravity=negative since opposes motion=mg
F and d are in opposite directions so  = 180 so:
W=Fgcos d=-Fgd
At the moment the ball reaches the highest point,
v=0 so K=0
Wtotal=K and K=1/2mv2
W=K0-K1=0-1/2 mv2=0-1/2(0.06kg)(50m/s)2=75J
W=-Fgd d=-W/Fg d=-W/mg
d=75J/(0.06kg)(10m/s2)=125m
Work-Energy Problems
• Often an easier way to attack
kinematics problems when force is
involved, especially if object is at rest at
start or finish since then kinetic
energy=0
• Can solve for any piece of the puzzleK, W, d, v
Potential Energy=U
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Energy an object has due to its position or configurationstored energy that can be retrieved.
Ex- height on a wave gives U, pulling back the string on a
bow gives it U.
Gravitational
Potential
Energy: Ug
• Potential energy
due to position
relative to surface
of the earth
• Ug=mgh
• Unit = Joule
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Gravitational Potential Energy: Ug
and Work done by gravity
• Gravity can do + or work depending on
motion
• Path independentdepends on
height, not path
taken
• Wg=mgh
• Where h is height
above arbitrary 0 pt
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Examples
• Physicsman (mass=60kg) scales a 40m
tall rock face. What is his potential
energy (relative to the ground)?
• Ug=mgh=(60kg)(10N/kg)(40m)=24000J
Take it 1 step further…
• If physicsman were to jump of the cliff,
what would his velocity be when he hits
the ground? Think…U is transformed
to K
• U…K…1/2mv2
• 24000J=1/2(60kg)v2
• V=28m/s
Work by Conservative vs.
Nonconservative Forces
• Conservative forces
are path
independent
– Ex: gravity
• Nonconservative
forces depend on
path
– Ex: kinetic frictionlonger path means
more work
If path is a closed loop, conservative forces W=0 but
nonconservative forces do work.
Ex. of conservative vs.
nonconservative
• In a roller coaster doing a closed
loop, the work done by gravity
(conservative) is - on way up and
+ on way down and cancels out by
the time it returns to start.
• Work done by kinetic friction
(nonconservative) is - (opposing
motion) the entire trip so doesn’t
cancel out when returns to start
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Work and Energy
• E=K+U
• E=1/2mv2+mgh
– Object’s mechanical energy is sum of kinetic and potential
energies
– Since U is relative to position, so is E
• Wnc=K+U
– Work done by nonconservative forces is sum of changes in
K and U
Conservation of Energy
• Since E=K+U, if no nonconservative forces (friction
for example) act on a system then mechanical
energy is conserved
• Ei=Ef or Ki+Ui=Kf+Uf
• If nonconervative forces act, then E is cot conservedenergy transformed into other (nonmechanical forms)
• In this case Ei+Wnc=Ef
• Think of what happens to your hands if you rub them
together producing friction
Ex: conservation of energy
• A ball of mass 2kg is gently pushed off
the lab table, 5.0m above the floor.
Find the speed of the ball as it strikes
the floor
• Ei=Ef or Ki+Ui=Kf+Uf
• 0+mgh=1/2mv2+0
• v=sqrt(2gh)=10m/s
Now try one with angles :)
• A skier starts from rest at the top of a 20
incline and skis a straight line to the bottom
of the slope 400m down the slope. If the
coefficient of kinetic friction between the
skis and the snow is 0.2, calculate the
skiers speed at the bottom of the run.
• Hint- draw a picture and use conservation
of mechanical energy- there is work done
by a nonconservative force!
Skier solved
Fk= kFN= kmgcos
Wnc=Wfriction=Fd= -( kmgcos )d
Ei+Wnc=Ef
0+mgh+ -( kmgcos )d=1/2mv2+0
h=dsin
mg(dsin ) -( kmgcos )d=1/2mv2
gd(sin - kcos )=1/2v2
v=sqrt[2gd(sin - kcos )]
=35m/s
• Power= rate at which work
gets done= work over time
• P=W/t
• Since W=Fd then P=Fd/t
and d/t=v
• P=Fv
• Unit= J/s=watt (W)
• Careful not to confuse unit
W (watt) with concept W
(work)
Power
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Ex: Power
• A mover pushes a large crate
mass=75kg across the truck bed for a
total distance of 6m. He exerts a
steady force of 300N for 20s. What is
his power output?
• P=W/t P=Fd/t=(300N)(6m)/20s=90W
Ex 2: Power
• What is the power output of an elevator
motor that can lift 1000kg at a constant
speed of 9.0m/s? Hint- what is the
force?
• F=mg since acting against weight
• P=W/t=Fd/t=Fv
• P=mgv=(1000kg)(10N/kg)(8.0m/s)
• =80000W=80kW