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Before we begin… enjoy the break. Before we begin… enjoy the break.

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Unit 4: Conservation of EnergyUnit 4: Conservation of Energy

Definition of Work (7-1, 7-2, 7-3)Definition of Work (7-1, 7-2, 7-3)• Examples of Work, Definition of Examples of Work, Definition of

Energy and the link to Work (7-3, 7-4, Energy and the link to Work (7-3, 7-4, 7-5)7-5)

• Potential Energy and Conservation of Potential Energy and Conservation of Energy (8-1, 8-2)Energy (8-1, 8-2)

• Problem Solving with the Problem Solving with the Conservation of Energy (8-3, 8-4)Conservation of Energy (8-3, 8-4)

• Applications of Conservation of Applications of Conservation of Energy (8-5,8-6, 8-7, 8-8, 8-9).Energy (8-5,8-6, 8-7, 8-8, 8-9).

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The General Expression for WorkThe General Expression for Work

• A precise definition of work is given by the A precise definition of work is given by the integralintegral

which is the area under which is the area under the curve in the Fcosthe curve in the Fcos vs l plane vs l plane

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

b

a

dlFW cos

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Work as a line integralWork as a line integral

b

a

b

a

ldFdlFW

cos

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Example 1: Work to Stretch a Example 1: Work to Stretch a SpringSpring• A person pulls on a spring A person pulls on a spring

stretching it from the stretching it from the equilibrium point to x=3.0cm. equilibrium point to x=3.0cm. The maximum force applied The maximum force applied (at the end of the stretch) is (at the end of the stretch) is 75N. How much work does 75N. How much work does the person do?the person do?

• We discussed this in SHM. To We discussed this in SHM. To stretch a spring, subject to stretch a spring, subject to Hooke’s law, the person must Hooke’s law, the person must apply a force F=+kx. apply a force F=+kx.

• The positive sign is required The positive sign is required since the person is pulling, the since the person is pulling, the spring is resisting with –kx.spring is resisting with –kx.

• But note already that the F But note already that the F and displacement are parallel, and displacement are parallel, so W=(Fcosso W=(Fcos)dx=(F)dx=(kx)dx)dx=(F)dx=(kx)dx

• We want the area under the We want the area under the kx versus x curve.kx versus x curve.

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The area of the triangle is just (1/2)bh. So The area of the triangle is just (1/2)bh. So w=(1/2)xkx = (1/2)kxw=(1/2)xkx = (1/2)kx22..

=Fcos=Fcos90=F

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77

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Example 2: Force as a Function of Example 2: Force as a Function of x.x.• A robot controls the A robot controls the

position of a camera with position of a camera with a servo motor that a servo motor that applies a force to a push-applies a force to a push-rod. The force is given rod. The force is given by:by:

• If the rod moves from If the rod moves from xx11=0.010m to =0.010m to xx22=0.050m how much =0.050m how much work did the robot do?work did the robot do?

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W=(2.5N)(0.04m)+0.5(15N)(0.04m) W=0.1J+0.3J~0.4J

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The Concept of Kinetic EnergyThe Concept of Kinetic Energy

• There are a number of forms of energy, energy of There are a number of forms of energy, energy of motion, energy stored in a field, the energy of heat….motion, energy stored in a field, the energy of heat….

• We start with the energy of motion or kinetic energy We start with the energy of motion or kinetic energy as it is easily related to the work done on an object.as it is easily related to the work done on an object.

• A moving object can do work on another object by A moving object can do work on another object by applying a force over a distance. Examples: applying a force over a distance. Examples: – A hammer striking a nail. It hits with great force and drives A hammer striking a nail. It hits with great force and drives

the nail a distance into the wall. the nail a distance into the wall. – The engine in a bus. It applies a force to the bus over the The engine in a bus. It applies a force to the bus over the

distance traveled.distance traveled.

• Note in both cases that the force resulted in a change Note in both cases that the force resulted in a change in the object’s final velocity. That is, the work led to in the object’s final velocity. That is, the work led to motion or kinetic energy. Let’s try to be more motion or kinetic energy. Let’s try to be more precise:precise:

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Kinetic EnergyKinetic Energy

• Consider an object of mass m Consider an object of mass m moving initially with speed vmoving initially with speed v11. .

• Now accelerate it uniformly to Now accelerate it uniformly to speed vspeed v22 by applying a constant by applying a constant force F over a distance d.force F over a distance d.

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The Work-Energy PrincipleThe Work-Energy Principle

• The connection between work and changing The connection between work and changing kinetic energy is a specific example of the work-kinetic energy is a specific example of the work-energy principle.energy principle.

• The ingredients were only the definition of work The ingredients were only the definition of work and the 2and the 2ndnd Law which applies to a net force. Law which applies to a net force.

• Note: Note: – If positive work is done the kinetic energy and velocity If positive work is done the kinetic energy and velocity

increase.increase.– Likewise a negative work decreases the kinetic energy Likewise a negative work decreases the kinetic energy

and velocity.and velocity.

• Kinetic energy goesKinetic energy goes– Linearly with massLinearly with mass– As the square of velocityAs the square of velocity

• Has units of Joules. Has units of Joules.

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1313

Hammer: Negative Work Slows DownNail: Positive Work Speeds up

Note we also have conservation of Energy

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Revisiting the Skier.Revisiting the Skier.• A 58-kg skier is coasting down A 58-kg skier is coasting down

a 25 degree slope. A kinetic a 25 degree slope. A kinetic frictional force of magnitude frictional force of magnitude FFfrfr = 70 N opposes his motion. = 70 N opposes his motion. At the top of the slope the At the top of the slope the skier’s speed is Vskier’s speed is Voo=3.6 m/s. =3.6 m/s. Determine the speed VDetermine the speed Vff at at point 57 meters down hill.point 57 meters down hill.

• Solution 1: Solution 1: We could sum the We could sum the forces in the x and y directions forces in the x and y directions and use Newton’s 2nd law to and use Newton’s 2nd law to set them equal to ma. After set them equal to ma. After solving for a we would have solving for a we would have the initial velocity, the initial velocity, displacement and displacement and acceleration. The problem acceleration. The problem could be completed with could be completed with application of the eqs. of application of the eqs. of motion. motion.

• Solution 2: Use the Work-Solution 2: Use the Work-Energy Theorem! Energy Theorem!

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• There are three forces:There are three forces:– normal forcenormal force– weightweight– kinetic frictional force. kinetic frictional force.

• Note that the normal Note that the normal force is balanced by the force is balanced by the component of the skier’s component of the skier’s weight perpendicular to weight perpendicular to the slope, since no the slope, since no acceleration occurs in acceleration occurs in that direction. that direction.

• There is, however, a net There is, however, a net force in the direction as force in the direction as the displacement. That’s the displacement. That’s the sum of forces in the the sum of forces in the x direction. x direction.

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Revisiting the Spring.Revisiting the Spring.

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

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

• But we also know that for But we also know that for a compressed spring a compressed spring W=(1/2)kxW=(1/2)kx22. Solving for k. Solving for k

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CommentsComments

Let’s take a moment to point out some Let’s take a moment to point out some important qualitative aspects of work: important qualitative aspects of work: 1)1) It is very important to remember that the It is very important to remember that the

work-energy theorem describes the net work-energy theorem describes the net external force. The theorem does not apply to external force. The theorem does not apply to the work done by an individual force (unless it the work done by an individual force (unless it is the only and net force.) is the only and net force.)

2)2) Also positive work increases kinetic energy or Also positive work increases kinetic energy or velocity and velocity and

3)3) Negative work decreases kinetic energy or Negative work decreases kinetic energy or velocity. velocity.

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2020

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A Simple Example: A PitchA Simple Example: A Pitch

• A 0.145kg baseball is thrown with a speed of A 0.145kg baseball is thrown with a speed of 25m/s.25m/s.– What is the kinetic energyWhat is the kinetic energy– How much work was done as the speed of the ball How much work was done as the speed of the ball

went from 0 to 25m/s?went from 0 to 25m/s?

• The last result shows that the precise path The last result shows that the precise path does not matter, just the initial and final does not matter, just the initial and final velocity.velocity.

• K(final)=(1/2)mvK(final)=(1/2)mv22=(1/2)(0.145kg)(25m/s)=(1/2)(0.145kg)(25m/s)22= = 45J45J

• Since W=KSince W=K22-K-K11=45J-0=45J.=45J-0=45J.

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2222

Another ExampleAnother Example

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

• From the definition From the definition of work we knowof work we know– WWnetnet=Fdcos(180)=-Fd=Fdcos(180)=-Fd– WWnetnet==K=0–(1/2)mvK=0–(1/2)mv22..

• Equating these last Equating these last two results: two results: – Fd=(1/2)mvFd=(1/2)mv22

• Rearranging: d=[0.5m/F]Rearranging: d=[0.5m/F](v(v22)=constant(v)=constant(v22))

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

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

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A SpringA Spring

• The spring at right has The spring at right has a spring constant of a spring constant of 360N/m.360N/m.– How much work is How much work is

required to compress it required to compress it from x=0 to x=11.0cm?from x=0 to x=11.0cm?

– If a 1.85kg block is placed If a 1.85kg block is placed against the spring and against the spring and the spring is released, the spring is released, what will be the speed of what will be the speed of the block when it leaves the block when it leaves the spring at x=0?the spring at x=0?

– What if it there is a What if it there is a coefficient of kinetic coefficient of kinetic friction equal to 0.38?friction equal to 0.38?

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2424• From our first problem:From our first problem:

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

• According to the work-According to the work-energy principle the block energy principle the block then gains that much in then gains that much in kinetic energy. K=(1/2)mvkinetic energy. K=(1/2)mv22 which can be solved for v: which can be solved for v:

• The frictional force is just The frictional force is just given by given by

• And the work is then justAnd the work is then just

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

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

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Kinetic Energy at Relativistic Kinetic Energy at Relativistic SpeedSpeed

• Einstein’s Theory of Einstein’s Theory of Relativity gives a Relativity gives a different expression different expression for kinetic energyfor kinetic energy

.103 toequal and

light of speed theis c where

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• The equationThe equation

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

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

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

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

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Quiz 2Quiz 2

• Average 26Average 26• DistributionDistribution

– Min 2Min 2– Max 50Max 50– 00-10: 600-10: 6– 11-20: 3011-20: 30– 21-30: 3321-30: 33– 31-40: 2531-40: 25– 51-50: 1251-50: 12

• It you need to see your point status please It you need to see your point status please see me after class or in my office.see me after class or in my office.