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Work and Energy

Potential EnergyWork is the measure of energy Work is the measure of energy transfer .transfer .

Objectives: After completing this unit, you should be able to:

• Define kinetic energy and potential energy, along with the appropriate units.

• Describe the relationship between work and kinetic energy, and apply the WORK-ENERGY THEOREM.

• Define and apply the concept of POWER.

EnergyEnergyEnergyEnergy is anything that can be is anything that can be converted into work; i.e., anything converted into work; i.e., anything that can exert a force through a that can exert a force through a distancedistance. The SI unit for energy is the joule (J)

Energy is the capability for doing Energy is the capability for doing work.work.

Potential EnergyPotential Energy:Potential Energy: Ability to do work Ability to do work by virtue of position or conditionby virtue of position or condition.

The SI unit for energy is the joule (J)

2

.

elevation above an arbitrary point

1 .

2

An elevated object U mgh

h

A stretched or compressed spring U kx

Potential EnergyPotential energy is measured as a difference from an arbitrarily chosen reference point

Example Problem: What is the potential energy of a 50-kg person in a skyscraper if he is 480 m

above the street below?

Gravitational Potential Gravitational Potential EnergyEnergy

What is the P.E. of a 50-kg person at a height of 480 m?

U = mgh = (50 kg)(9.8 m/s2)(480 m)

U = 235 kJU = 235 kJ

Kinetic EnergyKinetic Energy: Ability to do work Kinetic Energy: Ability to do work by virtue of motion. (Mass with by virtue of motion. (Mass with velocity)velocity)

Examples – Examples –

-A speeding carA speeding car

-A bulletA bullet

2

For an object with mass and moving at velocity

1Kinetic energy = =

2

m v

K mv

Examples of Kinetic Energy

2 21 12 2 (1000 kg)(14.1 m/s)K mv

What is the kinetic energy of a 5-g What is the kinetic energy of a 5-g bullet traveling at 200 m/s?bullet traveling at 200 m/s?

What is the kinetic energy of a What is the kinetic energy of a 1000-kg car traveling at 14.1 m/s? 1000-kg car traveling at 14.1 m/s?

5 g5 g

200 200 m/sm/s

2 21 12 2 (0.005 kg)(200 m/s)K mv 2

21(0.005 ) 200 1.00 10

2

mK kg x J

s

241

(1000 ) 14.1 9.94 102

mK kg x J

s

Work and Kinetic EnergyA force acting on an object changes A force acting on an object changes its velocity, and does work on that its velocity, and does work on that

object.object.

m

vi

m

vfd

F F

( ) ;W ork Fx ma x 2 2

0

2fv v

ax

2 21 12 2f iWork mv mv

The Work-Energy TheoremWork is Work is equal to the equal to the change inchange in ½mv½mv22

If we define If we define kinetic energykinetic energy as as ½mv½mv22 then we can state a very important then we can state a very important

physical principle:physical principle:

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a force is equal to the done by a force is equal to the change in kinetic energy that it change in kinetic energy that it produces.produces.

2 21 12 2f iWork mv mv

Example 1: A 20 g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find

the stopping force F if the entrance velocity is 80 m/s.

d

F = ?F = ?

80 80 m/sm/s

6 cm6 cm

Work = ½Work = ½ mvmvff2 2 - ½- ½ mvmvii

22 0

F d= - F d= - ½½ mvmvii22

FF (0.06 m) cos 180 (0.06 m) cos 18000 = - = - ½½ (0.02 kg)(80 (0.02 kg)(80 m/s)m/s)22

FF (0.06 m)(-1) = -64 J (0.06 m)(-1) = -64 J F = 1067 NF = 1067 N

Work to stop bullet = change in K.E. for Work to stop bullet = change in K.E. for bulletbullet

Example 2: A bus slams on brakes to avoid an accident. The tread marks of the tires are 25m long.

If k = 0.7, what was the speed before applying brakes?

d=25d=25 mm ffF= F= k.k.FFNN= = k k mgmg

Work =Work = F(F(cos cos ) ) dd

Work = - Work = - k k mg d mg d K =K = ½½ mvmvf f 2 2 - ½- ½ mvmvi i

22

-½-½ mvmvi i 22 = - = -k k mgmgd vvii= = 22kkgdgd

vi = 2(0.7)(9.8 m/s2)(25 m) vi = 18.52 m/s

vi = 18.52 m/s

Work = K

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a force is equal to the change in done by a force is equal to the change in kinetic energy that it produces.kinetic energy that it produces.

The Work-Energy Theorem:The Work-Energy Theorem: The work The work done by a force is equal to the change in done by a force is equal to the change in kinetic energy that it produces.kinetic energy that it produces.

Summary

Kinetic Energy:Kinetic Energy: Ability to do work Ability to do work by virtue of motion. (Mass with by virtue of motion. (Mass with velocity)velocity)

212K mv

Work = ½ mvf2 - ½ mvi

2Work = ½ mvf2 - ½ mvi

2

Potential Energy:Potential Energy: Ability to do work by Ability to do work by virtue of position or conditionvirtue of position or condition. 21

2

U mgh U kx

Conservation of EnergyLaw of conservation of energy – the total energy of a system remains unchanged unless an outside force does work on the system, or the system does work on something outside the system.

Conservation of EnergyThe energy of a system can interconvert between potential and kinetic energy, but the sum of the two will always remain constant.

Conservation of Energy

21

2U mgh K mv

E U K


2

If the ball falls straight down or slides down an incline without friction

E before the fall or the slide begins. Where is the initial height of the ball.

1E ' + '

2Where the pr

total

total

E mgh h

E mgh m v

max max

imed quantities represent the height and speed of the ball

at any point during the fall or slide.

2 Where is the velocity of the ball immediately before it hits zero height.v gh v


Sample problem1 : A 5.0 kg ball is dropped from a height of 10.0 meters. a. What is its velocity when it is 8.0 meters high?b. What is its velocity immediately prior to hitting the ground?


Solution: a. Before it is dropped the ball only has gravitational potential energy since there is no motion.Therefore: E=U+K=U+0=mgh=(10)(5)(9.81)= 490.5J=U

Conservation of EnergySolution (continued): a.(continued). At 8.0 meters the ball’s potential energyis U=(8)(5)(9.81)=392.4JSince its total energy is conserved, E=U+K and K=E-UK=490.5-392.4=98.1J

2 2 98.11 2 so 6.26

2 5.0

K mbut K mv v

m s


Solution: b. Right before the ball hits, h=0, U=0, and E=K+0=490.5J

2 490.52 14.0 2

5.0

K mv gh

m s


Pendulum – the energy of a pendulum is also a result of work done against gravity. The pendulum is at equilibrium when the bob hangs straight down.As the bob is moved from equilibrium it rises and acquires potential energy.



Pendulum – the potential energy of a pendulum displaced at an angle from equilibrium isU=mgh = mgl(1-cosWhere l is the length of the pendulumThe energy of the system E isE=mgl(1-cosmaxWhere max is the angle at the maximum displacement from equilibrium.


As the pendulum swings it loses potential energyand gains kinetic energy. At the bottom of the swingU=0 and K=E

2

max

1(1 cos )

2(1 cos )

U mgl K mv

E U K mgl


Example:A pendulum is 2.0m long, and has a bob that has a mass of 2.0 kg. The bob is pulled from the equilibrium position, so that the angle with the vertical axis θ=10o, and released.a.What is the potential energy of the system before the bob is released?b. What is the speed of the bob as it goes through the equilibrium position?c. What is the speed of the bob when the angle with the vertical is 5.0o?


Example:a.What is the potential energy of the system before the bob is released?U=mgl(1-cosmaxcos10o)=0.59J


Example:b. What is the speed of the bob as it goes through the equilibrium position?At equilibrium, U=0, and E=K

21 2 2(0.59) 0.768

2 2

K mK mv v

m s


Example:c. What is the speed of the bob when the angle with the vertical is 5.0o?K=E-UU=mgl(1-coscos5.0o)=0.149J

K=E-U=0.59J-0.149J=.441J

2(0.441)0.664

2

mv

s


2

Energy of a spring-mass system:

The potential energy of a spring- mass system is equal to the work done on

1the system U=

2where k is the force constant, and x is the displacement from equilibrium

kx

Conservation of EnergyAfter the mass is released, the spring stretches or compresses the velocity of the mass changes and the energy of the system interconverts between potential and kinetic energy.

The maximum displacement from the equilibrium position is called the amplitude A.

Conservation of energy




2 2

2

1 1

2 21

2

U kx K mv

E U K kA

Conservation of EnergyExample:A system consists of a spring attached to a 1.0kg mass that slides on a frictionless horizontal surface. The spring has a force constant of 200N/m. a. If the spring is stretched to a distance of 0.20m from the equilibrium position, what is the potential energy ofthe system?b. What is the speed of the mass when it is going through the equilibrium position?c. What is the speed of the mass when it is at a distance of 0.1 m from the equilibrium position?

Conservation of EnergyExample:a. If the spring is stretched to a distance of 0.20m from the equilibrium position, what is the potential energy ofthe system? Potential energy = the work done on the system

221 1(200) .20 4.0

2 2U kA J E

Conservation of EnergyExample:b. What is the speed of the mass when it is going through the equilibrium position?

At the equilibrium position, U=0 and K=E

2 422.83

1

K mv

m s

Conservation of EnergyExample:c. What is the speed of the mass when it is at a distance of 0.1 m from the equilibrium position?At 0.1m from equilibrium E=U+K

221 1200 0.10 1

2 24 1 3

2 322.45

1

U kx J

K E U J

K mv

m s


During an elastic collision – when two moving objects collide, the kinetic energy of the system is the same before and after the collision.

During an inelastic collision – when two moving objects collide, the kinetic energy of the system is not the same before and after the collision.