ppt work energy

8/2/2019 PPT Work Energy

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THE WORK OF A FORCE, THE PRINCIPLE OF

WORK AND ENERGY & SYSTEMS OF PARTICLES

Todays Objectives:

Students will be able to:

1. Calculate the work of a force.2. Apply the principle of work and

energy to a particle or system of

particles.


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QUIZ

1. What is the work done by the force F?

A) F s B) F s

C) Zero D) None of the above. s

s1 s2

F

2. If a particle is moved from 1 to 2, the work done on theparticle by the force, FRwill be

A) B)

C) D)

2

1

s

ts

F ds 21

s

ts

F ds 2

1

s

ns

F ds2

1

s

ns

F ds


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APPLICATIONS

A roller coaster makes use of gravitational forces to assist the

cars in reaching high speeds in the valleys of the track.

How can we design the track (e.g., the height, h, and the radius

of curvature, ) to control the forces experienced by thepassengers?


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WORK AND ENERGY

Another equation for working kinetics problems involving

particles can be derived by integrating the equation of motion

(F = ma) with respect to displacement.

This principle is useful for solving problems that involve

force, velocity, and displacement. It can also be used to

explore the concept ofpower.

By substituting at = v (dv/ds) into Ft = mat, the result is

integrated to yield an equation known as theprinciple of work

and energy.

To use this principle, we must first understand how to

calculate the work of a force.


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WORK OF A FORCE (Section 14.1)

A force does workon a particle when the particle undergoes a

displacement along the line of action of the force.

Work is defined as theproduct offorceand displacement components acting in

the same direction. So, if the angle

between the force and displacement

vector is , the increment of work dU

done by the force is

dU = F ds cos

By using the definition of the dot product

and integrating, the total work can be

written asr2

r1

U1-2 = F dr


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WORK OF A FORCE

(continued)

Work ispositive if the force and the movement are in thesame direction. If they are opposing, then the work is

negative. If the force and the displacement directions are

perpendicular, the work is zero.

IfF is a function of position (a common

case) this becomes

s2

s1

F cos dsU1-2

If both F and are constant (F = Fc), this equation furthersimplifies to

U1-2 = Fc cos s2 - s1)


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WORK OF A WEIGHTThe work done by the gravitational force acting on a particle

(orweight of an object) can be calculated by using

The work of a weight is the

product of the magnitude of the

particles weight and its vertical

displacement.

Ify is upward, the work isnegative since the weight force

always acts downward.

y

W

r2

r1

U1-2 = F dr yWyyWWdykdzjdyidxjWy

y

r

r

2

1

2

1

12).()(

y1

y2


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WORK OF A SPRING FORCE

When stretched, a linear elastic spring

develops a force of magnitude Fs = ks, where

k is the spring stiffness and s is thedisplacement from the unstretched position.

If a particle is attached to the spring, the force Fs exerted on theparticle is opposite to that exerted on the spring. Thus, the work

done on the particle by the spring force will be negative or

U1-2

= [ 0.5 k (s2

)2 0.5 k (s1

)2 ] .

The work of the spring force moving from position s1

to position

s2 is

= 0.5 k (s2)2 0.5 k (s1)

2k s dsFs dsU1-2

s2

s1

s2

s1


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WORK OF A SPRING FORCE

r2

r1

U1-2 = F dr

2

)(

22)().()(

2

2

2

1

2

1

2

2

2

1

2

1

2

1

ssk

ksksdsksdsFkdzjdyidsiF

s

s

s

s

s

r

r

s

Note s here is measured

with respect to the

upstretched position.


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PRINCIPLE OF WORK AND ENERGY

(Section 14.2 & Section 14.3)

U1-2 is the work done by all the forces acting on the particle as itmoves from point 1 to point 2. Work can be either apositive or

negative scalar.

By integrating the equation of motion, Ft = mat = mv(dv/ds), theprinciple of work and energy can be written as

U1-2 = 0.5 m (v2)2 0.5 m (v1)2 or T1 + U1-2 = T2

T1 and T2 are the kinetic energies of the particle at the initial and final

position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)

2.

The kinetic energy is always apositive scalar(velocity is squared!).

So, the particles initial kinetic energy plus the work done by all the

forces acting on the particle as it moves from its initial to final position

is equal to the particles final kinetic energy.


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PRINCIPLE OF WORK AND ENERGY

(continued)

The principle of work and energy cannotbe used, in general, to

determine forces directed normal to the path, since these forces

do no work.

Note that the principle of work and energy (T1 + U1-2 = T2) isnot a vector equation! Each term results in a scalar value.

Both kinetic energy and work have the same units, that of

energy! In the SI system, the unit for energy is called ajoule (J),

where 1 J = 1 Nm. In the FPS system, units are ftlb.

The principle of work and energy can also be applied to a system

of particlesby summing the kinetic energies of all particles in the

system and the work due to all forces acting on the system.


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The case of a body sliding over a rough surface merits special

consideration.

This equation is satisfied if P = kN. However, we know fromexperience that friction generates heat, a form of energy that doesnot seem to be accounted for in this equation. It can be shown thatthe work term (kN)s representsboth the external workof the

friction force and the internal workthat is converted into heat.

WORK OF FRICTION CAUSED BY SLIDING

The principle of work and energy would be

applied as

0.5m (v)2 + P s (kN) s = 0.5m (v)2

Consider a block which is moving over a

rough surface. If the applied forcePjust

balances the resultant frictional force kN,a constant velocity v would be maintained.


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Given: When s = 0.6 m, the spring is

not stretched or compressed,

and the 10 kg block, which issubjected to a force of F=

100 N, has a speed of 5 m/s

down the smooth plane.

Find: The distance s when the block stops.

Plan: Since this problem involves forces, velocity and displacement,apply the principle of work and energy to determine s.

EXAMPLE


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S2

S1

Fs

W

FNy2

y1

sin()=( y1-y2)/( s2-s1)

y1-y2

y2-y1= -(s2-s1) sin()


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Apply the principle of work and energybetween position 1

(s = 0.6 m) and position 2 (s). Note that the normal force (N)

does no work since it is always perpendicular to the

displacement.

EXAMPLE

(continued)Solution:

T1 + U1-2 = T2

There is work done by three different forces;1) work of a the force F =100 N;

UF = 100 (s s1) = 100 (s 0.6)

2) work of the block weight= -Wy=-(s2-s1) sin()*-mg

UW = 10 (9.81) (s s1) sin 30 = 49.05 (s 0.6)

3) and, work of the spring force.

US

= - 0.5 (200) (s0.6)2 = -100 (s

0.6)2


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The work and energy equation will be

T1 + U1-2 = T2

0.5 (10) 52 + 100(s 0.6) + 49.05(s 0.6) 100(s 0.6)2 = 0

125 + 149.05(s 0.6) 100(s 0.6)2 = 0

EXAMPLE

(continued)

Solving for (s 0.6),(s 0.6) = {-149.05 (149.05

2 4(-100)125)0.5} / 2(-100)

Selecting the positive root, indicating a positive spring deflection,

(s 0.6) = 2.09 m

Therefore, s = 2.69 m


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CONSERVATIVE FORCES, POTENTIAL ENERGY

AND CONSERVATION OF ENERGY


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CONSERVATIVE FORCE

(Section 14.5)

A forceF is said to be conservative if the work done is

independent of the path followed by the force acting on a particle

as it moves from A to B. This also means that the work done bythe forceF in a closed path (i.e., from A to B and then back to A)

is zero.

Thus, we say the work is conserved.

rF 0 d

x

y

z

A

BF

The work done by a conservative

force depends only on the positionsof the particle, and is independent of

its velocity or acceleration.


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CONSERVATIVE FORCE (continued)

A more rigorous definition of a conservative force makes

use of a potential function (V ).

The conservative potential energy of a particle/system is

typically written using the potential function V.

There are two major components to V commonly encountered

in mechanical systems, the potential energy from gravity andthe potential energy from springs or other elastic elements.

Vtotal = Vgravity + Vsprings


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POTENTIAL ENERGY

Potential energy is a measure of the amount of work a

conservative force will do when a body changes position.

In general, for any conservative force system, we can define

the potential function (V) as a function of position. The work

done by conservative forces as the particle moves equals the

change in the value of the potential function (e.g., the sum of

Vgravity and Vsprings).

It is important to become familiar with the two types ofpotential energy and how to calculate their magnitudes.


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POTENTIAL ENERGY DUE TO GRAVITY

The potential function (formula) for a gravitational force, e.g.,

weight (W = mg), is the force multiplied by its elevation from a

datum. The datum can be defined at any convenient location.

Vg = W y

Vg ispositive if y is above the

datum and negative if y is

below the datum. Remember,

YOU get to set the datum.


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ELASTIC POTENTIAL ENERGY

Recall that the force of an elastic spring is F = ks.

Notice that the potential

function Ve always yields

positive energy.

k s22

1Ve

Ve (where e denotes an

elastic spring) has the distance

s raised to a power (the

result of an integration) or


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CONSERVATION OF ENERGY

(Section 14.6)

When a particle is acted upon by a system of conservative

forces, the work done by these forces is conserved and the

sum of kinetic energy and potential energy remainsconstant. In other words, as the particle moves, kinetic

energy is converted to potential energy and vice versa.

This principle is called the principle of conservation of

energy and is expressed as

2211 VTVT = Constant

T1 stands for the kinetic energy at state 1 and V1 is thepotential energy function for state 1. T2 and V2represent these energy states at state 2. Recall, the

kinetic energy is defined as T = mv2.


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EXAMPLE

Given: The 2 kg collar is moving down

with the velocity of 4 m/s at A.

The spring constant is 30 N/m. The

unstretched length of the spring is

1 m.

Find: The velocity of the collar when

s = 1 m.

Plan:

Apply the conservation of energy equation between A andC. Set the gravitational potential energy datum at point A

or point C (in this example, we choose point A).


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Similarly, the potential and kinetic energies at A will be

VA = 0.5 (30) (2 1)2, TA = 0.5 (2) 4

2

EXAMPLE

(continued)Solution:

Note that the potential energy at Chas two parts.

VC = (VC)e + (VC)gVC = 0.5 (30) (5 1)

2 2 (9.81) 1

The kinetic energy at C is

TC = 0.5 (2) v2

The energy conservation equation becomes TA + VA = TC + VC .[ 0.5(30) (5 1)2 2(9.81)1 ] + 0.5 (2) v2

= [0.5 (30) (2 1)2 ]+ 0.5 (2) 42

v = 5.26 m/s

Vg = W y

k s221Ve


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2211 VTWVT consernon

2211 VTVT

For a system of particles under conservative forces:

For a particle under conservative and non conservative

forces:

Important Notes:


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Given: When s = 0.6 m, the spring is

not stretched or compressed,

and the 10 kg block, which is

subjected to a force of F=

100 N, has a speed of 5 m/s

down the smooth plane.

Find: The distance s when the block stops.

EXAMPLE

mghkmvssFmv

VTWVT consernon

22

212

2

1

2211

2

1

2

1)(0

2

1


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S2

S1

y2y1

sin()=( y1-y2)/( s2-s1)

y1-y2

h=y1-y2= (s2-s1) sin()

Datum


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)sin()(21

2

1

)(02

1

12

2

12

2

2

12

2

1

ssmgsskmv

ssFmv

Below is the solution of the other approach in numbers:

T1 + U1-2 = T2

0.5 (10) 52 + 100(s 0.6) + 10 (9.81) (s s1) sin 30 100(s 0.6)

2 = 0

Both are the same.


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Example

Given: The 800 kg roller

coaster starts from

A with a speed of

3 m/s.

Note that only kinetic energy and potential energy due

to gravity are involved. Determine the velocity at B using the

equation of equilibrium and then apply the conservation of

energy equation to find minimum height h .

Find: The minimum height, h, of the hill so that the car

travels around inside loop at B without leaving the track. Alsofind the normal reaction on the car when the car is at C for this

height of A.

Plan:


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Example (continued)

1) Placing the datum at A:

TA + VA = TB + VB

0.5 (800) 32 + 0= 0.5 (800) (vB)

2 800(9.81) (h 20) (1)

Solution:

manmg

NB 0

=

Equation of motion applied at B:

2v

mmaF nn

10800 (9.81) = 800

(vB)2

vB

= 9.905 m/s

2) Find the required velocity of the coaster at B so it doesntleave the track.


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Example (continued)

Now using the energy conservation, eq. (1), the minimum h

can be determined.

0.5 (800) 32 + 0 = 0.5 (800) (9.905)2 800(9.81) (h 20)

h= 24.5 m

manmg

NC

=

NC = 16.8 kN

3) To find the normal reaction at C, we need vc.

TA + VA = TC + VC

0.5 (800) 32 + 0 = 0.5 (800) (vC)2 800(9.81) (24.5 14) VC = 14.66 m/s

Equation of motion applied at B:

2vmFn

7NC+800 (9.81) = 800

14.662


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ATTENTION QUIZ

If the pendulum is released from the

horizontal position, the velocity of its

bob in the vertical position is _____

A) 3.8 m/s. B) 6.9 m/s.

C) 14.7 m/s. D) 21 m/s.

ppt work energy

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