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Before we begin… enjoy the break. Unit 4: Conservation of Energy. Definition of Work (7-1, 7-2, 7-3) Examples of Work, Definition of Energy and the link to Work (7-3, 7-4, 7-5) Potential Energy and Conservation of Energy (8-1, 8-2) Problem Solving with the Conservation of Energy (8-3, 8-4)

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unit 4 conservation of energy
Unit 4: Conservation of Energy
  • Definition of Work (7-1, 7-2, 7-3)
  • Examples of Work, Definition of Energy and the link to Work (7-3, 7-4, 7-5)
  • Potential Energy and Conservation of Energy (8-1, 8-2)
  • Problem Solving with the Conservation of Energy (8-3, 8-4)
  • Applications of Conservation of Energy (8-5,8-6, 8-7, 8-8, 8-9).

Physics 253

the general expression for work
The General Expression for Work
  • A precise definition of work is given by the integral

which is the area under the curve in the Fcosq vs l plane

  • This can be solved graphically.
  • Or using the integral calculus.

Physics 253

example 1 work to stretch a spring
Example 1: Work to Stretch a Spring
  • A person pulls on a spring stretching it from the equilibrium point to x=3.0cm. The maximum force applied (at the end of the stretch) is 75N. How much work does the person do?
  • We discussed this in SHM. To stretch a spring, subject to Hooke’s law, the person must apply a force F=+kx.
  • The positive sign is required since the person is pulling, the spring is resisting with –kx.
  • But note already that the F and displacement are parallel, so W=(Fcosq)dx=(F)dx=(kx)dx
  • We want the area under the kx versus x curve.

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slide6
=Fcosq=Fcos90=F

The area of the triangle is just (1/2)bh. So w=(1/2)xkx = (1/2)kx2.

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example 2 force as a function of x
We use the two geometric areas

sketched to guess:

W=(2.5N)(0.04m)+0.5(15N)(0.04m)

W=0.1J+0.3J~0.4J

Example 2: Force as a Function of x.
  • A robot controls the position of a camera with a servo motor that applies a force to a push-rod. The force is given by:
  • If the rod moves from x1=0.010m to x2=0.050m how much work did the robot do?

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the concept of kinetic energy
The Concept of Kinetic Energy
  • There are a number of forms of energy, energy of motion, energy stored in a field, the energy of heat….
  • We start with the energy of motion or kinetic energy as it is easily related to the work done on an object.
  • A moving object can do work on another object by applying a force over a distance. Examples:
    • A hammer striking a nail. It hits with great force and drives the nail a distance into the wall.
    • The engine in a bus. It applies a force to the bus over the distance traveled.
  • Note in both cases that the force resulted in a change in the object’s final velocity. That is, the work led to motion or kinetic energy. Let’s try to be more precise:

Physics 253

kinetic energy
V1

V2

F

d

Kinetic Energy
  • Consider an object of mass m moving initially with speed v1.
  • Now accelerate it uniformly to speed v2 by applying a constant force F over a distance d.

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the work energy principle
The Work-Energy Principle
  • The connection between work and changing kinetic energy is a specific example of the work-energy principle.
  • The ingredients were only the definition of work and the 2nd Law which applies to a net force.
  • Note:
    • If positive work is done the kinetic energy and velocity increase.
    • Likewise a negative work decreases the kinetic energy and velocity.
  • Kinetic energy goes
    • Linearly with mass
    • As the square of velocity
  • Has units of Joules.

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slide13
Hammer: Negative Work  Slows Down

Nail: Positive Work  Speeds up

Note we also have conservation of Energy

Physics 253

revisiting the skier
Revisiting the Skier.
  • A 58-kg skier is coasting down a 25 degree slope. A kinetic frictional force of magnitude Ffr = 70 N opposes his motion. At the top of the slope the skier’s speed is Vo=3.6 m/s. Determine the speed Vf at point 57 meters down hill.
  • Solution 1: We could sum the forces in the x and y directions and use Newton’s 2nd law to set them equal to ma. After solving for a we would have the initial velocity, displacement and acceleration. The problem could be completed with application of the eqs. of motion.
  • Solution 2: Use the Work-Energy Theorem!

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slide15
There are three forces:
    • normal force
    • weight
    • kinetic frictional force.
  • Note that the normal force is balanced by the component of the skier’s weight perpendicular to the slope, since no acceleration occurs in that direction.
  • There is, however, a net force in the direction as the displacement. That’s the sum of forces in the x direction.

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revisiting the spring
A 1200-kg car rolling on a horizontal surface with speed v=60km/hr strikes a coiled spring and is brought to rest in a distance of 2.2m. What is the spring constant of the spring?

We can use the change of kinetic energy to calculate the work done on the spring.

But we also know that for a compressed spring W=(1/2)kx2. Solving for k

Revisiting the Spring.

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comments
Comments

Let’s take a moment to point out some important qualitative aspects of work:

  • It is very important to remember that the work-energy theorem describes the net external force. The theorem does not apply to the work done by an individual force (unless it is the only and net force.)
  • Also positive work increases kinetic energy or velocity and
  • Negative work decreases kinetic energy or velocity.

Physics 253

a simple example a pitch
A Simple Example: A Pitch
  • A 0.145kg baseball is thrown with a speed of 25m/s.
    • What is the kinetic energy
    • How much work was done as the speed of the ball went from 0 to 25m/s?
  • The last result shows that the precise path does not matter, just the initial and final velocity.
  • K(final)=(1/2)mv2=(1/2)(0.145kg)(25m/s)2= 45J
  • Since W=K2-K1=45J-0=45J.

Physics 253

another example
An automobile is traveling at 60km/hr and can break to a stop in 20m with a constant breaking force. How long will it take to stop if moving at 120km/hr?

From the definition of work we know

Wnet=Fdcos(180)=-Fd

Wnet=DK=0–(1/2)mv2.

Equating these last two results:

Fd=(1/2)mv2

Rearranging: d=[0.5m/F](v2)=constant(v2)

In other words the distance is proportional to the velocity squared.

So if 60km/hr took 20m, then 120km/hr takes 80m.

Another Example

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a spring
A Spring
  • The spring at right has a spring constant of 360N/m.
    • How much work is required to compress it from x=0 to x=11.0cm?
    • If a 1.85kg block is placed against the spring and the spring is released, what will be the speed of the block when it leaves the spring at x=0?
    • What if it there is a coefficient of kinetic friction equal to 0.38?

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slide24
From our first problem:

That’s the energy stored in the spring, since by reversing the calculation we can show it’s also the work it provides when it pushes against the mass.

According to the work-energy principle the block then gains that much in kinetic energy. K=(1/2)mv2 which can be solved for v:

The frictional force is just given by

And the work is then just

The minus sign is required because the frictional force and displacement are in opposite directions.

The net work is then 2.18J-0.76J=1.42J. Which can be converted to velocity

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kinetic energy at relativistic speed
Kinetic Energy at Relativistic Speed
  • Einstein’s Theory of Relativity gives a different expression for kinetic energy

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slide26
The equation

Has been shown to be true for particles near the speed of light.

Notice that v cannot equal c or we would have infinite K.E. energy, a physical impossibility.

If v exceeded c we would have a negative energy- another physical impossibility.

Both of these observations are consistent with the fact that c represents a cosmic speed limit and nothing can obtain speeds faster than light.

Physics 253

quiz 2
Quiz 2
  • Average 26
  • Distribution
    • Min 2
    • Max 50
    • 00-10: 6
    • 11-20: 30
    • 21-30: 33
    • 31-40: 25
    • 51-50: 12
  • It you need to see your point status please see me after class or in my office.

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