Unit 4: Conservation of Energy

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Unit 4 Conservation of Energy

• This is the halfway point! Chapters 1- 6 covered
  kinematics and dynamics of both linear and
  rotational motion.
• Chapters 7-12 focus on the conservation laws of
  energy, momentum, and angular momentum and their
  applications.
• Although we cant get into details, the
  conservation laws are deeply associated with
  symmetries of nature.
• Symmetry wrt to time leads to energy conservation
• Symmetry wrt to position leads to momentum
  conservation.
• Well see that the application of the
  conservation laws provides a new way to
  understand and solve problems of motion.
• To start, we need to define work and then well
  look into the definition of energy.

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The Definition of Work

• Merriam-Websters
• 1 activity in which one exerts strength or
  faculties to do or perform something
• Followed by 10 more definitions!
• In physics, work has an exact definition
  associated with the a force as it acts on an
  object over a distance.
• To be precise work is defined to be the product
  of the magnitude of the displacement times the
  force parallel to the displacement.
• When you think about it the colloquial definition
  and the quantitative definition are not totally
  inconsistent. If you push that couch a distance
  along the floor youve applied a force the entire
  distance, and it sure feels like work!

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• F is the force
• d is the displacement
• q is the angle between the force and the
  displacement
• Note work is a scalar quantity
• Unit is N-m Joule(J)

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Limiting Cases

• Holding a 20kg mass object
• Standing Still
• The force you apply equals mg196N
• But d0
• WFdcosq F(0)0
• Walking forward at constant velocity
• Force 196N
• d is nonzero
• But cosq cos90 0
• WFdcosq Fd(0)0

• Motion and force in the same direction
• q 0 and cosq1
• WFd
• Example pushing the couch with 500N for 2m. The
  work would be 1000 Joules

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Example 1 Work on a Crate

• A 50-kg crate is pulled 40m by a 100N-force
  acting at a 37o angle. The force of friction is
  50N
• Determine the work done by the pulling,
  frictional, and net forces.

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• Pick x along the displacement vector.
• The free body diagram shows 4 forces
• FGmg50kg9.8m/s2490N
• FN490N
• Fp100N (given)
• Ffr50N (given)
• And the work
• WG(FG)(40m)cos(90)0
• WN(FN)(40m)cos(90)0
• WP(100N)(40m)cos(37)3200J
• WF(50N)(40m)cos(180)-2000N
• Note that the force pulling the mass does
  positive work and the force of friction does
  negative work.
• The net work is he sum or 1200J

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Example 2 Work on a Backpack

• A backpacker carries a 15.0 -kg pack up an
  inclined hill of 10.0m height.
• What is the work done by gravity and the net
  force on the backpack?
• Assume the hiker moves at a constant velocity up
  the hill.

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• There are two forces on the pack
• FG mg 147N
• The force of the hiker holding the pack aloft FH
  mg 147N
• The work done by the hiker is
• WH FH(d)(cosq) FHh 147N(10.0m) 1470J
• Only the height matters -not the distance
  traveled.
• The work done by gravity is
• WG FG(d)(cos(180-q)) FG(d)(-cosq)-FG(d)(
  cosq)-FGh -147N(10.0m)-1470J
• Once again only the height matters -not the
  distance traveled.
• The net work is just the sum of the work done by
  both forces or 0.

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A Technical Note The Scalar Product

• Although work is a scalar it involves two
  vectors force and displacement
• There is a mathematical operation called the
  scalar product that is very useful for the
  manipulation of vectors required to calculate
  work.
• The scalar or dot product is defined as
• Work can then be rewritten as

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Properties of the Scalar Product

• Since A, B, and cosq are scalars the scalar
  product is commutative
• The scalar product is also distributive
• If we define AAxiAyjAzk, and BBxi ByjBzk,
  then

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Example 3 Using the Dot Product

• A boy pulls a wagon 100m with a force of 20N at
  an angle of 30 degrees with respect to the
  ground.
• How much work has been done on the wagon?

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• With the axes shown
• FPFxiFyj (FPcosq)i (FPsinq)j17Ni10Nj
• d100mi
• Accordingly

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What about Varying Forces?

• So far weve only considered the work of a
  constant force.
• But its for more common to have a force varying
  with position
• A traveling rocket subject to diminishing gravity
• A simple harmonic oscillator
• A car with uneven acceleration
• We could break the motion into small enough
  intervals so that the work is more or less
  constant during each interval and then sum the
  work of the segments.
• Basically this is the idea behind an integral.

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• Consider an object traveling in the xy plane and
  subject to a varying force.
• We could just break the trajectory into small
  enough intervals such that the force is
  more-or-less constant during each interval.
• So for any particular interval labeled i
• And the total work for seven intervals would be

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• The sum can be graphically represented if
• Each shaded rectangle is one element of the sum.
• The curve represents the function Fcosq.
• Then, the work approximates the area under the
  curve.

• As we make Dli smaller and smaller, the sum of
  rectangles gives a better and better estimate of
  the area under the curve.
• In fact as it approaches zero we get an exact
  result for the area and for the work

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In closing

• The precise definition of work is given by
• Next well do some examples
• Youll see next it can also be interpreted as the
  amount of energy given to an object.
• Which opens the door to the conservation of
  energy Test WednesdaySee you Friday!