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Unit 4: Conservation of EnergyPowerPoint Presentation

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 can’t 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.

- We’ll 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 we’ll look into the definition of energy.

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The Definition of Work

- Merriam-Webster’s:
- 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 you’ve 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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Motion and force in the same direction

q = 0 and cosq=1

W=Fd

Example pushing the couch with 500N for 2m. The work would be 1000 Joules

Holding a 20kg mass object:

Standing Still:

The force you apply equals mg=196N

But d=0

W=Fdcosq = F(0)=0

Walking forward at constant velocity:

Force = 196N

d is nonzero

But cosq = cos90 = 0

W=Fdcosq = Fd(0)=0

Limiting CasesPhysics 253

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.

Example 1: Work on a CratePhysics 253

Pick x along the displacement vector.

The free body diagram shows 4 forces:

FG=mg=50kg*9.8m/s2=490N

FN=490N

Fp=100N (given)

Ffr=50N (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 A=Axi+Ayj+Azk, and B=Bxi+ Byj+Bzk, 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
- FP=Fxi+Fyj= (FPcosq)i+ (FPsinq)j=17Ni+10Nj
- d=100mi

- Accordingly

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What about Varying Forces?

- So far we’ve only considered the work of a constant force.
- But it’s 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 a varying force.

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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Physics 253 a varying force.

Physics 253 a varying force.

Physics 253 a varying force.

In closing a varying force.

- The precise definition of work is given by
- Next we’ll do some examples
- You’ll 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 Wednesday…See you Friday!

Physics 253

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