Work and Power

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Work and Power
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WORK

• Work is done when a force is exerted on an object
  causing the object to move in the direction of a
  component of the applied force.
• Work is a SCALAR quantity.
• The amount of work W F?d
• Where F applied force (in N)
• ?d distance the object moved ( in meters)

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WORK

• Or, it can be said that work is force through a
  distance.
• A force of one Newton acting through a distance
  of one meter is equal to one Joule (J) of work.
• (1 N)(1 m) 1 J (Joule is the SI unit of work)

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WORK
F
Fcos?
?d
In this example, the force doing work on the
crate would be Fcos? because that is the
magnitude of the force in the direction of motion.
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WORK

• A person carrying a box at a constant height is
  doing NO work on the box. F is perpendicular tot
  he direction of motion of the box, so F 0 and
  F?d also equals 0.
• Specify work done on object or by object and if
  work is by one force or a resultant force.

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OTHER UNITS FOR WORK

• System F?d Name of Unit
• cgs dynecentimeter erg
• British poundfoot foot-pound
• (ftlb)
• 1 J 107 erg 0.7376 ftlb

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Work Done by a Varying Force

• Examples
• 1. As a rocket moves away from the earth, work
  is done to overcome the force of gravity, which
  varies as the inverse square of the distance
  from the earths center.
• 2. Force exerted by a spring is not constant as
  it becoes fully extended of fully compressed.

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Work Done by a Varying Force
Force in the direction of motion (N)
?d
?d
1
3
5
Distance (m)
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Work in Rotary Motion

• Work can be done in rotary motion
• Linear Work F?d
• Rotary Work Fr ??
• To change linear to rotary work ?d r ??
• Combining with ? Fr
• the equation becomes W ? ??

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Example of Work in Rotary Motion
How much work is done in rotating a 4-m diameter
wheel through 1 revolution while using a force of
12 N?
W ? ?? Fr ?? ?? 2? W Fr 2? W (12 N)(2
m)(2)(3.14) W 150 J
r 2.0 m F 12 N ?? 1 rev
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POWER

• Power is the rate at which work id done, or the
  rate at which energy is transformed (such as in
  an internal combustion engine).
• Power (P)
• SI units P J/s Watts (W)
• British P ftlbs / s horsepower (hp)

Work(W) time (t)
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POWER

• P W / t (F?d) / ? t
• and ?d/? t velocity (v)
• so, P Fv

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James Watt

• The unit for power was first chosen by James Watt
  (1736-1819), who needed a way to specify the
  power of his newly developed steam engines. He
  found by experiment that a good horse can work
  all day at an average rate of about 360 ftlbs/s.
  So as not to be accused of exaggeration in the
  sale of his steam engines, he multiplied this by
  1 1/2 when he defined the horsepower.
• 1 hp 550 ftlbs/s 746 W

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Power in Rotary Motion

• P W/t
• and W ? ?? (in radians)
• So, P ? ?? / t
• and ?? / t ? (omega)
• ? P ? ? (torque x angular velocity)

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Example of Power

• A 220 kg box was pushed across a level floor with
  a force of 20 N in 12 seconds. What amount of
  power was necessary in (a) Watts and (b)
  horsepower?

Given m 220 kg ?d 7 m F 20 N t 12 seconds
P W / t W F?d (20 N)(7 m) 140 J P 140 J
/ 12 s 11.67 W
1 hp 746 W
0.015 hp
11.67 W x
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Example Power in Rotary Motion

• A car engine can apply 150 Nm of torque to the
  wheels of a car that then spin at a rate of 120
  rad/s. What is the power of the engine in (a)
  Watts and (b) horsepower?

P ? ? P (150 Nm)(120 rad/s) 18,000 W
Given ? 150 Nm ? 120 rad/s
1 hp 746 W
18,000 W x
24.13 hp
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SIMPLE MACHINES
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SIX SIMPLE MACHINES
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LEVER
FUP
FDOWN
fulcrum
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LEVER
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Mechanical Advantage
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Mechanical Advantage

• OR Output Force Input Force
• From the earlier example
• Output force 15 N
• Input force 5 N
• Ratio is 155 or 31 mechanical advantage of 3

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Ideal Mechanical Advantage

• Ideal mechanical advantage (IMA)
• Ratio of effort distance to load distance
• Effort distance/load distance 1 1/3 3
• If there is no friction, IMA MA
• IMA is only for cases without friction!

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Actual Mechanical Advantage

• Actual Mechanical Advantage (AMA)
• AMA is the ratio of load force to effort force
• Load force/effort force 15N/5N 3
• Use AMA when friction is present!

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Types of Levers

• Type 1 Lever Fulcrum between force and load
• Increase force at expense of distance
• Ex seesaw, crowbar

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Types of Levers

• Type 2 Load is between fulcrum and input force
• Increase force at expense of distance
• EX hand bottle cap opener, wheelbarrow

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Types of Levers

• Type 3 Fulcrum at one end, load at other
• Increase distance at expense of force
• EX elbow, construction crane

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PULLEYS

• Machine that can change the direction of a force.
  Pulleys in combinations can also multiply forces

output
input
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PULLEYS

• The single pulley acts like a type 1 lever where
  the axis of the pulley is the fulcrum
• - both sides (radii) are the same son only the
  direction of the force is changed
• - no forces are multiplied

output
input
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PULLEYS

• This single pulley acts like a type 2 lever
• - the load is between the fulcrum and input
  force
• - each strand of rope supports 1/2 the load, so
  the MA is 2

MA of pulley system of strands of rope
supporting the load This system both changes
direction of force multiplies input force
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EFFICIENCY

• Efficiency is the ratio of the useful work output
  to the total work input.
• Efficiency Work Out Work In
• This number will ALWAYS be less than 100
• Lower efficiency means more energy lost
• Most lost energy is heat

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ENERGY

• Mechanical energy is the ability to do work
• this is not true for all energy
• Two types of mechanical energy
• 1. Kinetic - energy of motion
• 2. Potential - stored energy

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KINETIC ENERGY

• An object in motion has the ability to do work
  and can be said to have energy.
• When an object has kinetic energy and does work,
  the two are related by
• Work ?KE of the object
• Or KE ½ mv2

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KINETIC ENERGY

• This quantity is the translational kinetic energy
  of the object.
• The net work done on an object is equal to its
  change in kinetic energy. This is known as the
  work-energy principle.
• Kinetic energy is a scalar quantity and is
  measured in Joules (J).
• The kinetic energy of a group of objects is just
  the algebraic sum of the individual kinetic
  energies of all the objects.
• KEtotal ? ½ m1v12 ½ m2v22 ½ m3v32

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EXAMPLES

• A 145 g baseball is thrown with a speed of 25
  m/s. (a) What is its KE (b) How much work was
  done to reach this speed starting from rest?
• KE ½ mv2 ½ (.145 kg)(25 m/s)2 45 J
• Since the initial KE 0, then W 45 J
• How much work is required to accelerate a 1000 kg
  car from 20 m/s to 30 m/s?
• W ?KE ½ mv22 ½ mv12
• W ½ (1000kg)(30m/s)2 ½ (1000kg)(20m/s)2
• W 2.5 x 105 J

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POTENTIAL ENERGY

• Potential energy can be in many forms such as
  mechanical, chemical, gravitational, and elastic.
• wound up clock spring
• calories in food
• raising an object above the earth
• pulling back on a slingshot
• Most common form is gravitational potential

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POTENTIAL ENERGY

• EX A heavy brick held high in the air has
  potential energy because of its position relative
  to the earth. If released, it has the ability to
  do work on a stake, driving it into the ground.
• Work F?d mgh
• m mass
• g gravity
• h height above reference point (usually the
  earth)

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POTENTIAL ENERGY

• It requires a certain amount of work to raise the
  object up, so the object has a potential to do
  work (PE).
• PE mgh
• The amount of PE depends on the vertical height
  of the object above some reference level.
• PE is path independent.
• it does not matter what path it took to get to
  that height, just the fact that it is up there

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EXAMPLE

• A 1000 kg roller coaster car moves form point A
  to point B and then to point C.
• What is its gravitational PE at points B and C
  relative to point A?
• What is the change in PE when it goes from B to
  C?

B
A
10 m
5 m
C
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• (a) Call point A y 0 (or h 0)
• At point B PEB mghB (1000)(9.8)(10)
  980,000J
• At point C PEC mghC (1000)(9.8)(-5) -
  490,000J
• or, 1.5 million Joules less than at point A
• (b) PEC PEB 980,000 490,000 1.7 x 105J
• The PE decreases by 2.5 x 105 J.

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CONSERVATION OF MECHANICAL ENERGY

• The total mechanical energy of any system is the
  sum of all the potential and kinetic energies.
• E ?KE ?PE
• Remember, energy can NEVER be created or
  destroyed, only transformed from one form to
  another.
• So, as KE increases PE decreases and vice versa.

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CONSERVATION OF MECHANICAL ENERGY

• When energy does appear to be lost, look for
  other places it went. (Friction, heat, sound,
  deformation, etc.)
• THE TOTAL ENERGY IN ALL SYSTEMS WILL REMAIN
  CONSTANT.

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CONSERVATION OF MECHANICAL ENERGY

• Wile E. Coyote is once again chasing the
  roadrunner. HE misses a turn and runs over the
  edge of a 200 m cliff. Just as in the cartoons,
  he pauses for a second and then begins to fall.
  If he has a mass of 25 kg, find the following
• )PE and KE at the top of the cliff
• )PE and KE at the bottom of the cliff
• )PE and KE midway through the fall
• )How fast he is moving just before he hits the
  ground
• NOTE Neglect air resistance

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• (A)At the top, there is no downward velocity,
• so v0 and KE 0 J
• The PE would be mgh or
• PE (25)(9.8)(200)
• PE 49,000 J
• (B) At the bottom, the height would be 0, so PE
  0
• So, with the Law of Conservation of
  Energy,
• At the top PE KE 49,000J
• At the bottom PE KE 49,000J
• So that means KE 49,000J

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• (C)Midway through the fall, PE mgh
  (25)(9.8)(100) 24,500 J
• Or exactly ½ the total energy in the system
• And E PE KE 49,000 24,500 KE
• so KE 24,500J
• (d) At the bottom, KE 49,000J
• KE ½ mv2 49,000
• 49,000 ½ (25) v2
• v2 3920 m2/s2
• v 62.61 m/s or about 140 mph

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ROTATIONAL KINETIC ENERGY

• A body rotating about an axis has rotational
  kinetic energy.
• Linear and rotational KE are found in the same
  way.
• LINEAR ROTARY
• KE ½ mv2 KE ½ I?2
• m mass I moment of inertia
• v velocity ? angular velocity

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ROTATIONAL KINETIC ENERGY

• An object that rotates while its center of mass
  undergoes translational motion will have both
  translational and rotational KE.
• EXAMPLES
• Wheel of a car
• Ball rolling down a hill
• KEtotal ½ mvcm2 ½ Icm?2
• cm stands for center of mass

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ELASTIC POTENTIAL ENERGY

• In general, the change in potential energy
  associated with a particular force, when an
  object is moved from one position to another, is
  found to be the work that would be done by that
  force in moving the object from the second back
  to the first.
• Examples of Elastic PE
• Rubber Band
• Coiled Spring
• Slinky!

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ELASTIC POTENTIAL ENERGY

• It takes a certain amount of force to elongate or
  compress a spring. As the original length of the
  spring is changed further (?d or ?x) a larger
  force (F) is necessary. This force changes for
  different materials (brass vs. iron), so we use a
  proportionality constant, k.
• K is the spring constant and is a measure of the
  stiffness of the particular spring. It is
  measured in N/m.

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ELASTIC POTENTIAL ENERGY

• The force the spring exerts is in the opposite
  direction, so
• F -kx
• This is called the Spring Equation, or Hookes
  Law.
• This equation will hold true as long as ?x is not
  too great, and it shows the restoring force.

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Calculating the PE of a compressed spring

• Find the work needed to compress it.
• W Fx
• x change in distance from the original
  position
• F force necessary to cause the change
• This force varies, so you are finding the average
  force
• Finitial 0 (not compressed at all)
• Ffinal F (fully compressed)

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• Favg (Ffinal - Finitial) / 2 ½ (F-0) ½ F
• And F kx
• So, Favg ½ kx
• W ( ½ kx)(x) ½ kx2
• And PE W
• So, PE ½ kx2

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• If a spring is stretched instead of compressed,
  it still follows these rules. However, if you
  exceed the springs elastic limit, then it will
  be permanently destroyed.
• REMEMBER
• 1. F is positive if we exert a force on the
  spring
• 2. F is negative if the spring exerts a force on
  an object

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Example

• It takes 4 J of work to elongate a spring 0.5 m.
  What is the spring constant of this particular
  spring?
• GIVEN PE ½ kx2 W
• W 4 J 4 J ½ k(0.5)2
• ?x 0.5 m 4 0.125k
• k 32 N/m

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LINEAR MOMENTUM

• Linear momentum is defined as the product of an
  objects mass and its velocity.
• p mv
• p momentum
• m mass
• v velocity
• Since velocity is a vector, so is momentum. It
  is measured in kilogram-meters per second
  (kgm/s).
• The more momentum an object has, the harder it is
  to stop.

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EXAMPLES

• A small, fast moving car vs. large, slow moving
  bus
• A fast moving Mack truck vs slow moving
  motorcycle
• A force is required to change the momentum of an
  object, whether it is to increase the momentum,
  decrease the momentum, or to change its
  direction.

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• Newton originally stated his 2nd Law in terms of
  momentum
• The rate of change of momentum of a body is
  proportional to the net force applied to it.
• ?F ?p/?t
• ?F is the vector sum of all the forces acting on
  a body
• ?p is the resulting momentum change during the
  time interval ?t
• As with energy, momentum is always conserved

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CONSERVATION OF MOMENTUM

• The total momentum of an isolated system of
  bodies remains constant.
• System a set of objects that interact with each
  other
• Isolated system one in which the only forces
  present are those between the objects of the
  system
• The sum of these forces is zero due to Newtons
  3rd Law
• If there are external forces present, Newtons
  3rd Law does not hold true and momentum is not
  conserved.

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COLLISIONS AND IMPULSE

• Collisions
• Baseball and bat
• Two billiard balls colliding
• Two cars colliding
• Atomic particles Bubble chamber
• When two or more objects collide, all objects
  become temporarily OR permanently deformed due to
  the large initial force. The initial force
  increases from zero to F in milliseconds. This
  jump in force occurs over a distinct period of
  time, ?t.

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• From Newtons 2nd Law, F ?p/?t
• Or ?p F?t
• Which states that the change in momentum of an
  object is equal to the amount of force applied to
  an object in the time, ?t.
• The quantity F?t is called the impulse.
• Impulse change in momentum ?p
• Any time a force (or net force) acts on an
  object, momentum is changed.
• We can use this in our momentum equations

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COLLISIONS

• Two Types
• Elastic a collision in which the total KE of a
• system is conserved
• Inelastic KE is not conserved energy is
  changed to thermal, potential, energy of
  deformation, etc.

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Elastic Collisions in One Dimension

• When two solid objects strike each other, no heat
  is produced (nor deformation), so KE is
  conserved. It also follows that the momentum
  (p) is conserved.
• This is an ideal case since there will always be
  a minimal amount of lost energy. (Sound, heat,
  etc.)
• EXAMPLE
• Two billiard balls, one at rest and one moving at
  0.4 m/s collide.

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• v 0.4 m/s v 0 m/s
• We will call the original velocities v1 and v2
  before the collision and v1 and v2 after the
  collision.
• So, from the conservation of momentum
• m1v1 m2v2 m1v1 m2v2
• And, because it is an elastic collision, KE is
  conserved.
• ½ m1v1 ½ m2v2 ½ m1v1 ½ m2v2

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• You can use either equation to find what is being
  asked for you can also use the two to solve
  simultaneous equations.
• You can find m1 , v1 , m2 , v2 , v1, or v2
• Combining the two equations gives us the result
• v1 - v2 v2 - v1
• So, for any elastic head-on collision, the
  relative speed of the two particles after the
  collision is the same as before the collision.

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INELASTIC COLLISIONS

• A collision that does not conserve KE is called
  an inelastic collision.
• one or both objects deformed or the objects stick
  together
• If the two objects stick together, it is called a
  completely inelastic collision.
• Two railroad cars joining
• Two balls of clay hitting
• Even though the KE is not conserved, the total
  energy and momentum are.

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• Using the two railroad cars as an example
• A runaway car (mass of 10,000 kg) moving at 24
  m/s strikes another car at rest (same mass) and
  the two become joined. How fast does the pair
  move after the collision?

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BEFORE
AFTER
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• GIVEN
• m1 10,000 kg
• m2 10,000 kg
• v1 24 m/s
• v2 0 m/s

BEFORE p m1v1 m2v2 AFTER p
(m1m2)v m1v1 m2v2 (m1m2)v (10000)(24)
(10000)(0) 20000 v 240,000 0 20,000
v v 12 m/s
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• We find that the two move off with ½ the speed of
  the first. Now, see if this is an elastic or
  inelastic collision.
• KE BEFORE ½ m1v12 ½ m2v22
• ½ (10000)(24)2 ½ (10000)(0)2
• KE 2.88 x 106 J
• KE AFTER ½ (m1 m2) v2
• ½ (20,000)(12)2
• KE 1.44 x 106 J
• The KE after is ½ of the original, so KE is not
  conserved this is an inelastic collision.

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CONSERVATION OF ANGULAR MOMENTUM

• As with KE, p can be changed to rotational
  motion.
• Linear momentum p mv
• Angular momentum L I?
• Where L angular momentum
• I moment of inertia
• ? angular velocity
• the total angular momentum of a rotating body
  remains constant if the net torque acting on it
  is zero.

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EXAMPLE

• A skater doing a spin on ice rotates at a
  relatively low speed when her arms are
  outstretched. When she brings her arms in close
  to her body, she suddenly spins much faster.
• WHY moment of inertia I ?mr2
• When she brings her arms in, I decreases since L
  is conserved (LI?), her angular velocity has to
  increase to keep a constant, so she spins faster.