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AAST/AEDT

AP PHYSICS B:

WORK IN PHYSICS

By definition the mechanical work or simply work is expressed as a product of the

applied force times displacement times cos , where is the angle between the directions

of the displacement and the force.

W = F Dcos

cos appears because only horizontal component of the force is responsible for the

motion and it can be expressed as F cos

The unit of work is 1 Joule.

It equals to 1 Newton times

1 meter.

The mechanical work can

be positive, negative and

even it can be equal to zero.

Example: An object is moving along the plane surface as shown on a diagram

The pulling force produces positive work (the angle between the force and the

displacement is zero and the cos 0 =1)

The friction force produces negative work (the angle between the force and the

displacement is 180° and the cos 180° =-1)

The normal force and weight does not produce any work (the angle between the

force and the displacement is 90° and the cos 90° =0)

If a constant force is

applied on the object, than the

force vs. displacement graph

will be

The force is equal to the

width of the rectangle and the

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displacement is equal to its length. Thus, if we multiply one by another the product would

be the area.

CONCLUSION: On the force vs. displacement graph the work is numerically equal

to the area.

The law allows us to derive a formula for the work done by a variable force. For

example if we compress a spring from the equilibrium position to the compression of x,

we need to apply the force that equals kx, where k is the elastic coefficient (spring

constant).

This equation F = kx, is called Hooke’s law.

The force vs. displacement graph for that force

is presented on a diagram.

We have proved above that work is

equal to the area. Thus to determine the work

done by the elastic force, we need to estimate

the area of the shaded triangle.

The formula is valid only for the work done by an elastic force, when the final

deformation is zero.

If elastic force does work from the

initial deformation of X to the final

deformation of Xf, than the area on the

graph would be computed as the area of a

trapezoid.

Another example. We want to express

the work done by the Gravity force when the

object drops from the height of h.

In the model the force is equal to mg

and the displacement is h.

Thus, we have

W= mgh.

It is important to note that the work done by

the gravity force depends only on the total

W kxx

2or W

kx2

2

W kXf kX

2 X Xf

k(X2 X f

2)

2

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height and does not depend on the trajectory.

We can easily prove that

Let us assume that an object is

moving along the curved trajectory.

We can present that trajectory as an

infinite broken line that consists only of

the horizontal and the vertical lines

Work done by the gravity force on a

vertical segments is equal to mghi,

where i is the number of the segment.

On a horizontal segments the work

done by gravity is zero, because the

force is perpendicular to the displacement. The total work done is equal to

W = mgh1 + mgh2 +mgh3 +... = mg (h1 + h2 + h3 +...) = mgh,

where h is the total height

We have proved that for the elastic and for the gravity forces work done depends

only on the final and initial positions of the object and it is independent on the shape

of the path. The forces with that property are defined as conservative forces.

Another example of the conservative forces is electromagnetic forces.

For Nonconservative forces the work done depends on the path. Friction and air

resistance are the examples. Sometimes they are defined as dissipative forces, because

when they are present - the mechanical energy dissipates into the heat.

Power

Let us assume that you have to go to the second floor of your house from the basement.

That means that you have to move 5 m up. Your mass is 70 kg. The work you need to

perform is mgh and it is equal to 70 kg*9.8 m/s25 m = 3430 J.

First time you move very slowly and it takes you 30 sec to reach the second floor. Next

time you are in a rush, and you do it in 10 seconds.

Both times you do equal mechanical work. The only difference is the rate of

performance.

The physical quantity that shows us the rate of doing work is defined as POWER. It is a

scalar quantity

Power equals to the work done divided by the time taken to make this work

In SI units power is measured in Joules/s. That unit is called Watt. We have

1 W = 1 J/s

P W

t1

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I remind you, that 1 kilowatt = 103 W, 1 Megawatt = 10

6 W. These units are often used

in engineering.

One more unit is often used. It is horsepower 1 hp = 736 W

There is a second way to express power. If we substitute the formula for work W=F.D

into (1) the result is

Power equals to the applied force times velocity.

MECHANICAL ENERGY

Let us run an experiment. We observe two equal balls with a mass of m.

The first one is located at a height of h above the ground while the second is at the

ground level. We hold both of them. If we

release the second ball nothing happens. If

we release the first one, it drops and does

the mechanical work.

Thus, one object has the ability to do work

and the second one has not. That object’s

property to be able to do work is defined

as energy. If the object is able to do work it

possesses the energy.

Now we have to define the method to

compute energy. Let us consider an object

above the Earth.

It is obvious that our object is able

to do small amount of work when it falls a short distance. It does more work when the

distance is greater.

It produces the maximum mechanical

work, when it travels the maximum

distance.

For the value of energy we assign the

maximum mechanical work that an

object can perform under the given

conditions.

From that definition it is easy to

understand, that the unit of energy, has

to be the same as the unit of

mechanical work, i.e. Joule (J).

KINETIC ENERGY

Let us assume that an object of mass m is moving with a velocity of V.

P FD

t FV 2

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It hits the block, exerts a

force on it and starts to do

work on it. So, the object

has the ability to do work.

It possesses energy.

The energy that an

Object possesses because

of its motion is defined as kinetic energy (KE);

Of course when object does work its velocity decreases. The maximum work will be

done, if its final velocity becomes zero. By the definition we know that the energy is

equal to the maximum work done. Thus, we have

where V- is the objects initial velocity, Vf is the final velocity.

The formula

is widely used in physics. The formula is an expression for the kinetic energy of the

object.

WORK-ENERGY THEOREM

There is an interesting relationship between the mechanical work done by the object and

its kinetic energy. We run the same experiment as the earlier one, but that time let us

assume that the ball

changes its velocity

from V1 to V2.

Then the work done:

If we eliminate all intermediate results we obtain

The net mechanical work done on an object is equal to the change in its kinetic

energy.

It easy to realize that if work done on an object is positive, than the object’s energy

increases and vice versa.

KE Wmax FD maV

2Vf

2

2amV2

2

KE mV 2

2

W FD maV2

2 V1

2

2amV2

2

2mV1

2

2 KE2 KE1 KE

W KE

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POTENTIAL ENERGY

By definition

POTENTIAL ENERGY IS THE ENERGY ASSOCIATED WITH THE

INTERACTIONS BETWEEN THE DIFFERENT OBJECTS OR WITH THE

INTERACTIONS BETWEEN THE DIFFERENT PARTS OF THE ONE AND

THE SAME OBJECT.

Examples:

1.

An object at a certain height above the Earth

possesses mechanical energy (i.e. it is able to

perform echanical work) because of its interaction

with the Earth. (because of the existence of the

gravitational force).

2. Let us assume that we have two balls, charged with opposite charges.

They attract one another with the electrical forces. If we release one ball, it starts

to move and does mechanical work. Thus, it

possesses mechanical energy because of the

interaction with the other ball.

3. Let us assume, that we have a magnet and a

steel ball. The ball is attracted to the magnet and being released it moves to the magnet

and does mechanical work. Thus, it possesses mechanical energy because of its

interaction with the magnet.

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4. The stretched spring is able to compress and to do the mechanical work because of the

interactions between the different parts of

the stretched spring

All these examples are the examples of the

objects with a potential energy.

We will study potential energy in different

parts of physics. Today we will study only

the mechanical potential energy.

POTENTIAL ENERGY OF THE OBJECT ABOVE THE EARTH

(Gravitational PE)

Let us again assume that an object is positioned above the Earth at a height of h. (see. the

diagram above).

As we know the energy is equal to the maximum work that the object is able to do. It is

obvious that the maximum work will be done when our ball reaches the ground and won't

be able to do any more work. Thus, we have.

PE = Wmax = mgh,

where h is the initial height of the object.

POTENTIAL ENERGY OF THE SPRING

(Elastic PE) Let us assume we have a stretched spring with a deformation x. When spring contracts,

the maximum work will be done when the final deformation is 0. Thus, we have

WORK -POTENTIAL ENERGY RELATIONSHIP

Let us run an experiment. The ball is falling from the height h1 to the

height h2 and the distance traveled is h

So, we have

W = mgh = mg(h1- h2) = mgh1- mgh2

We do know, that mgh1 is the PE of the object at height h1 and that the

mgh2 is the PE of the object at height of h2.

We continue:

W = mgh1- mgh2= PE1-PE2= - ( PE2-PE1)= -∆PE

PE Wmax kx2

2

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because to compute ∆PE we have to subtract initial value of PE from the final and not

vice versa.

Our final result is

The net mechanical work done on an object is equal to its change in potential

energy with an opposite sign.

If you compare that result with a work-energy theorem you discover that the only

difference is in a sign of change, That can be easily understand.

Examples:

1. If a ball is falling down, the positive work done by the gravitational force on a

ball decreases PE (∆PE is negative) and increases KE (∆KE is positive)

2. If the ball is thrown up vertically, the work done by the gravitational force on a

ball is negative (force and displacement are opposite by the direction). PE does increase

(∆PE is positive) and the KE does decrease (∆KE is negative).

THE MAIN PROPERTY OF THE POTENTIAL ENERGY

Let us observe a ball in the room. If somebody asks you to compute the potential

energy of that ball, it is hard to give an answer. The problem is that we do not know what

height we have to plug into the formula of

PE. That means that the ball possesses one

energy respectively to the table, another one

respectively to the floor and the third one and

even negative respectively to the ceiling,

Thus, when we compute potential energy, it is

mandatory to specify respectively to what

initial level it is defined. Usually that is the

level of the ground.

MECHANICAL ENERGY CONSERVATION LAW

We will prove the law for one particular example. Let us assume

that the free falling ball with a mass of m initially was at the height

of h1 with a velocity of V1. After a certain time the ball would be at a

position of h2 with a velocity of V2.

The distance traveled is h. According to the formulas for the free

fall motion the height traveled is

From the diagram we can easily observe, that h = h1-h2. Thus,

h V2

2 V1

2

2g

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If we multiply both parts of the equation by mg, the result is

We can easily observe that

So, we have

If we move PE2 to the right side and KE1 to the left side of the equation, the result is

On the left side we have total mechanical energy at the first position. On the right side is

the total mechanical energy at the second position. We proved that the total mechanical

energy is conserved.

This law is valid only if the gravity or elastic forces (conservative forces) are

exerted on the object. If the friction force (nonconservative force) is applied, then

the total mechanical energy does not conserve. The part of the mechanical energy

transforms to another nonmechanical type of energy, for example into the heat

energy.

Energy Conservation Law (Solving Problems)

Energy conservation law is the most powerful tool that can be used for solving problems

in physics. The general strategy consists of the several steps. They are:

1. You have to understand what are the initial conditions of the physical system and what

is the total mechanical energy for it.

2. Write an expression for the total initial energy for the physical system on the right

side of the equation.

3. Try to realize what happen with the object's energy; i.e. how does the energy

transform?

Generally there are two possible ways for the energy transformation

a. It could be transformed from the one type of the mechanical energy into the

other, for example from PE to KE.

b. It could be completely or partially transformed into the work done against so

called dissipative forces (friction, air resistance, etc.). Finally as the result of this

work, the energy transforms from the mechanical type into the thermal one.

4. Writes the expression for the final mechanical energy + the work done against the

dissipative forces on the left side of equation.

h1 h2 V2

2 V1

2

2g

mgh1 mgh2 mgV2

2 V1

2

2gor mgh1 mgh2

mV2

2

2mV1

2

2

mgh1 PE1; mgh2 PE2;mV2

2

2 KE2;

mV1

2

2 KE1

PE1 PE2 KE2 KE1

PE1 KE1 PE2 KE2

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5. You have to obtain the energy conservation law for yours particular problem.

It has to be something like PE1+ KE1= PE2+KE2+W, but not exactly the same.

It can be different for the different problems.

6. Instead of expressions PE, KE, and W write the formulas for them.

7. Solve the obtained equation respectively to the unknown variable.

Example: The ball with a velocity of V is moving along the table. The kinetic friction

coefficient is µ. What is the stopping distance?

In the problem the initial KE of the ball finally transfers into the work done against the

friction force. Thus, the energy conservation law is KE =W, or

because the friction force F = µmg we have

After expressing the distance, we obtain the final formula

Example 2: Motion on the

vertical plane

The problem is: What is the

minimum value of initial H of the

ball, enough to travel through the

loop the loop device with a radius

of R?

At the top point A the ball

should experience centripetal

mV2

2 FD

mV2

2 mgD

D mV 2

2mg

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acceleration, provided by the gravity force.

That allows us to compute the minimum velocity that the ball should posses at the top

point. (we can cancel m and express v.)

Now we can apply the energy conservation law.

At the initial point B the ball possesses only potential energy. At point A it has both

kinetic and potential.

PEB= KEA+PEA

or

(the height at the point A is 2R.)

If we cancel m and substitute the value of V at point A, we have

Home assignment:

Cuttnell Chapter 6,

Pages 179-180: Conceptual Questions: #1-10

Pages 180-… Problems, 3,4,8,9,13, 15,20, 21, 23, 27, 29,33, 37,41,42, 46, 49, 52,

55, 59, 67, 76

mg mV2

R

V gR

mgH mV

2

2mg2R

gH gR

2 g2R or H

5R

2

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Problem #3

#4 1.88E+07 J

#8a 1804.21 J

b -1201.46 J

#20a 830.29 N

b 9133.19 J

c -8524.89 J

d 3.92 m/s 3.92428337

#52 a 3.12 J

b 23.97 N