Transcript Work, Power, & Efficiency

Work, Power, &
Efficiency
Work
• Work is done when a force is exerted
on an object and the object moves a
distance x.
Work = force ·distance
W =F· x
Unit: N·m = J (called a Joule)
Work is a scalar quantity.
Work
When you hold something, are you exerting a
force on the object? Yes.
When you hold something, are you doing
work? No.
If you set the object on a table, does the table
exert a force on the object? Yes. Does the
table do any work? No.
When you hold something, then, you are not
doing work either.
Work
Work
• Work is only done by a force on an
object if the force causes the object to
move in the direction of the force.
Work
• When a force is
applied at an angle,
use the component
of the force in the
direction of motion.
W  F  cos θ  Δx
Work
• For work done by the frictional force:
W  Ff  Δx
• The work is negative because friction
acts in a direction opposite to the
motion x.
Ff  μ  m  g
Work
• For an object moving at constant speed
(a = 0 m/s2), Fx = Ff.
• If the force and distance are parallel ( = 0°;
cos 0 = 1), the amount of work is positive,
but if the two vectors are anti-parallel ( =
180°; cos 180 = -1), then the work is
negative.
When the displacement
is in the same direction
as the force, the work
is positive.
When the displacement
is in the opposite
direction of the force,
the work is negative.
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Work Done by a Gravitational Force
• One constant force we already have
dealt with is the force of gravity.
• So we should be able to compute
the work done on an object as it
rises and falls…
• So lets look at a particle-like
tomato of mass m that is thrown
upward with an initial velocity v0.
• As the tomato rises, work is done
on the tomato by the force of
gravity.
Work Done by a Gravitational Force
• The work done is: W = m  g  d  cos
• Because the tomato is rising, the
force is opposite in direction to
the displacement ( = 180º;
cos 180 = -1), thus the work done
is negative.
W = -m  g  d
• As we expect, the tomato’s
velocity decreases until it reaches
zero; it then “turns around” and
begins to fall.
θ
Work Done by a Gravitational Force
• At this time, the force is in
the same direction as the
displacement  = 0º;
cos 0 = 1), so the work done
is positive.
W = m  g  d  cos θ
W = mg d
Work Done in Lifting
and Lowering an Object
• When we lift an object, our applied force
does positive work Wa on an object while at
the same time gravity is doing negative
work Wg on the object:
Wapplied = m  g  d
Wgravity = -m  g  d
• This also applies when an object is lowered.
Work Done in Lifting
and Lowering an Object
• So if the
displacement is
vertically upward,
the work done is:
Wapplied = m  g  d
• If the displacement
is vertically
downward, then
the work done is: W
gravity
= -m  g  d
Work Done in Lifting
and Lowering an Object
• Notice that the work done on the
object:
W = ±m  g  d
is independent of the magnitude of the
force exerted on the object.
Work and Inclined Planes
• Fx = gravitational component
• W = Fx·x
• The direction of motion is positive.
– Fx is negative if the object moves up the
incline.
– Fx is positive if the object
moves down the incline.
Fx  m  g  sin θ
Fx  Fw  sin θ
Work and Inclined Planes
• For masses on a horizontal surface or
on an incline, the work done by the
normal force Fn = 0 J.
• The normal force is not in the
direction of the motion, therefore,
x = 0 m.
Work and Inclined Planes
• Refer to the problem to determine if
you have to worry about friction. If
frictionless, set Ff = 0 N.
Elastic Objects/The Spring Force
• We know from common experience that the harder
you push on a spring, the harder it pushes back;
the same is true for pulling…
• The force needed to stretch or compress an elastic
object is a varying force, increasing with increasing
distance from the equilibrium position (the
unstretched position, where x = 0).
Spring (Elastic) Constant k
•
•
•
•
F = -k·x (this equation is called Hooke’s law)
x = distance of stretch/compression
k = spring constant; unit = N/m.
The spring constant is a measure of the stiffness of
the spring; the larger k is, the stiffer the spring is
(and the more force is needed for a given
displacement)
• Negative sign indicates that the force F of the spring
is in the opposite direction to the distorting force
(the pull or the push on the spring). In the problems
you will solve, the negative sign is often dropped.
• The force F is always directed toward the equilibrium
position.
• Work = 0.5·k·
x2
F
k
Δx
Springs in Parallel
• Each spring will
share the weight Fw
and stretch the
same distance x.
• Effective spring
constant = k1 + k2
Fw
k1  k 2 
Δx
Springs in Series
• Each spring supports the
entire weight Fw and will
stretch independently.
Fw
Fw
k1 
Δx1 
Δx1
k1
Fw
Fw
k2 
Δx 2 
Δx 2
k2
Springs in Series
• Total distance of stretch = x1 + x2
• Effective spring constant:
Fw
k
Δx1  Δx 2
Machines
1.
2.
3.
4.
5.
6.
Pulleys
Levers
Inclined planes
Wheel and axle
Wedge
Screw
• Lever Applet
• Pulley Applet
Machines
• Machines can change the direction of a
force and/or multiply the force.
• Work comes in two forms: input and
output
– Work input (Win) = Feffort·xeffort
– Work output (Wout) = Fresistance·xresistance
Machines
• Ex: push down on a lever to raise a load
Fw; the weight Fw is the resistance force.
• Work input (Win) is what you do; the effort
force Fe that you apply to the machine and
the distance this force moves xe.
Machines
• Work output (Wout) is what gets done;
the resistance force FR that is moved
by the machine and the distance this
force moves xR.
• Wout cannot be greater than Win due to
friction.
Machines
• Ex: Inclined planes require only a
force equal to the parallel component
of the weight (Fx), but this force has to
be applied over a longer distance L.
Win  Fe  L  Fx  L
Wout  FR  h  Fw  h
Mechanical Advantage
• Mechanical Advantage (MA): indicates the
amount by which the effort force Fe is
increased by the machine.
FR
MA 
Fe
Dimensionless
(no units)
• Ex: if the MA of a machine is 4, the
machine multiplies Fe by 4 to get FR; if Fe =
10 N, then FR = 4·10 N = 40 N.
• MA does take friction into account.
• The MA of a pulley is equal to the number
of ropes that support weight.
• There is a trade off
between force and
distance; if it takes
less force, the force
has to move
through a greater
distance.
Ideal Mechanical Advantage
• Ideal Mechanical Advantage (IMA):
the mechanical advantage of a
machine assuming 100% efficiency –
no loss of force to overcoming friction.
Dimensionless
xe
IMA 
(no units)
xR
Efficiency
• Efficiency: the ratio of work output to
work input.
Wout
e
 100 %
Win
FR  x R
e
 100 %
Fe  x e
MA
e
IMA
Fw  h
lifting : e 
 100 %
Fe  x e
Efficiency
• Efficiency is always less than 1 (as a
decimal) or 100% due to frictional
losses.
• The higher the efficiency, more of the
work input is converted to work
output; less of the work output is lost
to overcoming friction.
Power
• Power: the rate at which work is done.
W
P
t
• Unit: J/s = W (watt)
• 1 kilowatt (kW) = 1000 W
Power
• Work = force·distance
• Velocity = distance/time
W Fx
P

Fv
t
t
• Horsepower (hp): power produced by a
typical British working horse.
• 1 hp = 746 W
torque  rpm
• horsepower in a car engine: hp 
5252
Things To Look For
• Pulling an object with friction: flat surface
Fy  F  sin θ
Fx  F  cos θ
Fn  Fw  Fy
Ff  μ  Fn
• Work done by the pulling force F:
W = Fx·x
Work done by friction: W = -Ff·x
Things To Look For
• Pushing an object with friction: flat surface
Fy  F  sin θ
Fx  F  cos θ
Fn  Fw  Fy
Ff  μ  Fn
• Work done by the pushing force F:
W = Fx·x
Work done by friction: W = -Ff·x
Things To Look For
• Pushing an object up an incline with
friction: at constant speed (a = 0 m/s2)
Fn  Fy  Fw  cos θ
Fx  Fw  sin θ
Ff  μ  Fn
Fa  Fx  Ff
W  Fa  x
Things To Look For
• If there is an acceleration up the incline
W = (Fx + Ff + m·a)·x
• For an elevator problem with friction:
Fup = Fw + Ff = m·g + Ff
P = Fup·v
• If there is an acceleration:
Fup = Fw + Ff + m·a = m·g + Ff + m·a
P = Fup·v