Transcript work

Introduction to Work
Energy and Work
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A body experiences a change in energy when one
or more forces do work on it. A body must move
under the influence of a force or forces to say work
was done.
A force does positive work on a body when the
force and the displacement are at least partially
aligned. Maximum positive work is done when a
force and a displacement are in exactly the same
direction.
If a force causes no displacement, it does zero
work.
Forces can do negative work if they are pointed
opposite the direction of the displacement.
Calculating Work in Physics B
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If a force on an object is at least partially aligned
with the displacement of the object, positive work is
done by the force. The amount of work done
depends on the magnitude of the force, the
magnitude of the displacement, and the degree of
alignment.
W= F r cos q
F
F
q
q
r
Forces can do positive or negative work.
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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
Ranking Task 1
mg
Units of Work
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SI System:
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British System:
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foot-pound
cgs System:
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Joule (N m)
erg (dyne-cm)
Atomic Level:
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electron-Volt (eV)
Problem: A droplet of water of mass 50 mg falls at constant speed under
the influence of gravity and air resistance. After the drop has fallen 1.0
km, what is the work done by a) gravity and b) air resistance?
Problem: A sled loaded with bricks has a mass of 20.0 kg. It is pulled at
constant speed by a rope inclined at 25o above the horizontal, and it
moves a distance of 100 m on a horizontal surface. If the coefficient of
kinetic friction between the sled and the ground is 0.40, calculate
a)The tension in the rope.
b)The work done by the rope on the sled
c)The work done by friction on the sled.
Work and a Pulley System
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A pulley system, which has at least
one pulley attached to the load, can
be used to reduce the force
necessary to lift a load.
Amount of work done in lifting the
load is not changed.
The distance the force is applied over
is increased, thus the force is
reduced, since W = Fd.
F
m
Work as a “Dot Product”
Calculating Work a Different Way
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Work is a scalar resulting from the multiplication of
two vectors.
We say work is the “dot product” of force and
displacement.
W=F•r
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W= F r cos q
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dot product representation
useful if given magnitudes and directions of vectors
W = Fxrx + Fyry + Fzrz
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useful if given unit vectors
The “scalar product” of two vectors is
called the “dot product”
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The “dot product” is one way to multiply two
vectors. (The other way is called the “cross
product”.)
Applications of the dot product
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Work
Power
Magnetic Flux
W=Fd
P=Fv
ΦB = B  A
The quantities shown above are biggest
when the vectors are completely aligned and
there is a zero angle between them.
Why is work a dot product?
F
q
s
W=F•r
W = F r cos q
Only the component of force aligned
with displacement does work.
•Problem: Vector A has a magnitude of 8.0 and vector B has a magnitude of
12.0. The two vectors make an angle of 40o with each other. Find A•B.
•Problem: A force F = (5.0i + 6.0j – 2.0k)N acts on an object that undergoes
a displacement of r = (4.0i – 9.0j + 3.0k)m. How much work was done on the
object by the force?
• Problem: A force F = (5.0i – 3.0j) N acts upon a body which undergoes a
displacement d = (2.0i – j) m. How much work is performed, and what is
the angle between the vectors?
Work by Variable Forces
Work and Variable Forces
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For constant forces
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For variable forces, you can’t move far until
the force changes. The force is only
constant over an infinitesimal displacement.
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W=F•r
dW = F • dr
To calculate work for a larger displacement,
you have to take an integral
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W =  dW =  F • dr
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
• Problem: Determine the work done by the force as the
particle moves from x = 2 m to x = 8 m.
F (N)
40
20
0
-20
-40
2
4
6
8
10
12
x (m)
Problem: A force acting on a particle is Fx = (4x – x2)N. Find
the work done by the force on the particle when the particle
moves along the x-axis from x= 0 to x = 2.0 m.
• Problem: Derive an expression for the work done by a
spring as it is stretched from its equilibrium position
• Problem: How much work does an applied force do when it
stretches a nonlinear spring where the force varies according
to the expressions F = (300 N/m) x – (25 N/m2) x2 from its
equilibrium length to 20 cm?
Work Energy Theorem
Net Work or Total Work
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An object can be subject to many forces at the same
time, and if the object is moving, the work done by
each force can be individually determined.
At the same time one force does positive work on
the object, another force may be doing negative
work, and yet another force may be doing no work
at all.
The net work, or total, work done on the object (Wnet
or Wtot) is the scalar sum of the work done on an
object by all forces acting upon the object.
Wnet = ΣWi
The Work-Energy Theorem
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Wnet = ΔK
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When net work due to all forces acting upon an
object is positive, the kinetic energy of the object
will increase.
When net work due to all forces acting upon an
object is negative, the kinetic energy of the object
will decrease.
When there is no net work acting upon an object,
the kinetic energy of the object will be unchanged.
(Note this says nothing about the kinetic energy.)
Kinetic Energy
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Kinetic energy is one form of mechanical energy,
which is energy we can easily see and
characterize. Kinetic energy is due to the motion
of an object.
K = ½ m v2
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K: Kinetic Energy in Joules.
m: mass in kg
v: speed in m/s
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In vector form, K = ½ m v•v
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Ranking Tasks 1 2 3 4
Problem: A net force of 320 N acts over 1.3 m on a 0.4 kg
particle moving at 2.0 m/s. What is the speed of this
particle after this interaction?
• Problem: Calculate the kinetic energy change of a 3.0 kg object that
changes its velocity from (2.0 i + 2.0 j -1.0 k) m/s to (-1.0 i + 1.0 j 2.0 k) m/s. How much net work done on this object?
Problem: A force of F1 = (4.0 i + j) N and another of F2 = -4.0 j N
act upon a 1 kg object at rest at the origin. What is the speed of
the object after it has moved a distance of 3.0 m?
Power
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Power is the rate of which work is done.
No matter how fast we get up the stairs, our
work is the same.
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When we run upstairs, power demands on our
body are high.
When we walk upstairs, power demands on our
body are lower.
Pave = W / t
Pinst = dW/dt
P=F•v
Units of Power
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Watt = J/s
ft lb / s
horsepower
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550 ft lb / s
746 Watts
• Problem: A 1000-kg space probe lifts straight upward off the planet Zombie,
which is without an atmosphere, at a constant speed of 3.0 m/s. What is the
power expended by the probe’s engines? The acceleration due to gravity of
Zombie is ½ that of earth’s.
• Problem: Develop an expression for the power output of an airplane
cruising at constant speed v in level flight. Assume that the
aerodynamic drag force is given by FD = bv2.
• By what factor must the power be increased to increase airspeed by
25%?
How We Buy Energy…
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The kilowatt-hour is a commonly used unit by
the electrical power company.
Power companies charge you by the kilowatthour (kWh), but this not power, it is really
energy consumed.
• Problem: Using what you know about units, calculate how
many Joules is in a kilowatt-hour.
Workday
Conservative and NonConservative Forces
More about force types
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Conservative forces:
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Work in moving an object is path independent.
Work in moving an object along a closed path is zero.
Work is directly related to a negative change in potential
energy
Ex: gravity, electrostatic, magnetostatic, springs
Non-conservative forces:
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Work is path dependent.
Work along a closed path is NOT zero.
Work may be related to a change in mechanical energy, or
thermal energy
Ex: friction, drag, magnetodynamic
Potential Energy
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A type of mechanical energy possessed by an
object by virtue of its position or configuration.
Represented by the letter U.
Examples:
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Gravitational potential energy, Ug.
Electrical potential energy , Ue.
Spring potential energy , Us.
The work done by conservative forces is the
negative of the potential energy change.
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W = -ΔU
Gravitational Potential Energy (Ug)
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The change in gravitational potential
energy is the negative of the work done by
gravitational force on an object when it is
moved.
For objects near the earth’s surface, the
gravitational pull of the earth is roughly
constant, so the force necessary to lift an
object at constant velocity is equal to the
weight, so we can say
ΔUg = -Wg = mgh
Note that this means we have defined the
point at which Ug = 0, which we can do
arbitrarily in any given problem close to
the earth’s surface.
h
Fapp
mg
Spring Potential Energy, Us
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Springs obey Hooke’s Law.
Fs(x) = -kx
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Ws =  Fs(x)dx = -k  xdx
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Fs is restoring force exerted BY the spring.
Ws is the work done BY the spring.
U s = ½ k x2
Unlike gravitational potential energy, we
know where the zero potential energy point is
for a spring.
•Problem: Three identical springs (X, Y, and Z) are hung as
shown. When a 5.0-kg mass is hung on X, the mass
descends 4.0 cm from its initial point. When a 7.0-kg mass is
hung on Z, how far does the mass descend?
X
Y
Z
• Sample problem: Gravitational potential energy for a body a large distance
r from the center of the earth is defined as shown below. Derive this
equation from the Universal Law of Gravity.
Gm1m2
Ug  
r
• Hint 1: dW = F(r)•dr
• Hint 2: ΔU = -Wc (and gravity is conservative!)
• Hint 3: Ug is zero at infinite separation of the masses.
Conservation of Mechanical
Energy
System
Boundary
Law of Conservation of Energy
The system is isolated and boundary
allows no exchange with the
environment.
E = U + K + Eint
= Constant
No mass can enter or leave!
No energy can enter or leave!
Energy is constant, or conserved!
Law of Conservation of
Mechanical Energy
We only allow U
and K to
interchange.
We ignore Eint
(thermal energy)
E=U+K
= Constant
Law of Conservation of Mechanical
Energy
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E=U+K=C
for gravity
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E = U + K = 0
 Ug = mghf - mghi
 K = ½ mvf2 - ½ mvi2
(What assumptions are we making here?)
for springs
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or
 Us = ½ kxf2 - ½ kxi2
 K = ½ mvf2 - ½ mvi2
(What assumptions are we making here?)
Ranking Tasks 1
Pendulum Energy
h
½mv12 + mgh1 = ½mv22 + mgh2
For any points two points in the pendulum’s swing
Spring Energy
0
m
½ kx12 + ½ mv12
= ½ kx22 + ½ mv22
-x
For any two points in a spring’s
oscillation
m
m
x
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Problem: A single conservative force of F = (3i + 5j) N acts on a 4.0 kg particle.
Calculate the work done if the particle if the moves from the origin to r = (2i - 3j) m.
Does the result depend on path?
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What is the speed of the particle at r if the speed at the origin was 4.0 m/s?
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What is the change in potential energy of the system?
•Sample Problem: A bead slides
on the loop-the-loop shown. If it is
released from height h = 3.5 R,
what is the speed at point A? How
great is the normal force at A if the
mass is 5.0 g?
Non-conservative Forces and
Conservation of Energy
Non-conservative forces
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Non-conservative forces change the
mechanical energy of a system.
Examples: friction and drag
Wtot = Wnc + Wc = K
Wnc = K – Wc
Wnc = K + U
Sample Problem: A 2,000 kg car starts from rest and coasts down
from the top of a 5.00 m long driveway that is sloped at an angel of 20o
with the horizontal. If an average friction force of 4,000 N impedes the
motion of the car, find the speed of the car at the bottom of the
driveway.
• Problem: A parachutist of mass 50 kg jumps out of a hot air balloon
1,000 meters above the ground and lands on the ground with a speed of
5.00 m/s. How much energy was lost to friction during the descent?
Conservation of Energy Lab
Force and Potential Energy
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In order to discuss the relationships between
potential energy and force, we need to review
a couple of relationships.
Wc = Fx (if force is constant)
Wc =  Fdx = - dU = -U (if force varies)
 Fdx = - dU
Fdx = -dU
F = -dU/dx
Stable Equilibrium – 1st and 2nd
Derivatives
U
x
Unstable Equilibrium – 1st and 2nd
Derivatives
U
x
Neutral Equilibrium – 1st and 2nd
Derivatives
U
x
More on Potential Energy and Force
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In multiple dimensions, you can take
derivatives of each dimension separately.
F(r) = -dU(r)/dr
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Fx = -U/  x
Fy = -U/  y
Fz = -U/  z
F = Fxi + Fyj + Fzk
Molecular potential energy diagrams
U
R
• Problem: The potential energy of a two-particle system separated by a
distance r is given by U(r) = A/r, where A is a constant. Find the radial
force F that each particle exerts on the other.
• Problem: A potential energy function for a two-dimensional force is of the
form U = 3x3y – 7x. Find the force acting at a point (x,y).