Work

and
Energy Reported By
Micko C. Renomeron
&
Angelica Jones Antoya
Work and Energy

• One type of problem that frequently occurs in dynamics involves a

relation between force, displacement, and velocity. A technique
known as the work-energy method is particularly adapted for solving
such problems. As we shall see, this method eliminates consideration
of acceleration and leads directly to the desired solution.
Work and Energy

• The terms work and kinetic energy are used to define certain mathe-
matical expressions.
• A precedent for this was found when we defined the moment of
inertia of an area as equivalent to the mathematical expression
‫𝜌 ׬‬2 ⅆ𝐴.
Work and Energy
• In the following articles we shall do two things:
1. Derive the mathematical expressions which define work and
kinetic energy as applied to translation, rotation, and plane
motion
2. Discuss the technique, use, and advantages of the work-
energy method as applied to these several motions.
Work and Energy
Fundamental Work-Energy Equation for Rectilinear Translation
𝑤
σ𝑥 = 𝑎
𝑔 Equation A
a ds = v dv
Eliminating a in the two equations results in
𝑤
σ 𝑥 ⅆ𝑠 = 𝑣 ⅆ𝑣 Equation B
𝑔

Integrate,

𝑠 𝑤 𝑣
‫׬‬0 σ 𝑥 ⅆ𝑠 = ‫ 𝑣 𝑣׬‬ⅆ𝑣
𝑔 𝑜
Equation C

𝑠 1𝑊 1𝑊 𝑊
‫׬‬0 σ 𝑥 ⅆ𝑠 =
2 𝑔
v2 -
2 𝑔
v 0= (v2-
2
2𝑔
v20)
Work and Energy
Fundamental Work-Energy Equation for Rectilinear Translation
Work-Energy Equation for Constant Forces
1𝑊 1𝑊 𝑊
σ 𝑥 ⅆ𝑠 = v2 - v20= (v2- v20)
2 𝑔 2 𝑔 2𝑔

1𝑾 2
Substituting these units 𝑣
2𝒈

1 𝑾 2 𝒍𝒃 𝒇𝒕𝟐
Kinetic Energy = 𝑣 = x sec2 x 𝟐 = ft-lb
2𝒈 𝒇𝒕 𝒔𝒆𝒄
Work and Energy
Meaning of Work
Consider a body subjected to the constant forces
shown in the figure which move the body up the cline
σ 𝑥 = P cos 0 – W sin 0 – F (a)
σ 𝑥 . S = (P cos 0)s – (w sin 0)s – F . s (b)
Work and Energy
A. Application of the Work-Energy Method
B. Resultant Work. Variable Forces
C. Power. Efficiency
D. Work-Energy Applied to Curvilinear Translation
E. Fundamental Work-Energy Equation for Rotation
F. Work-Energy Method with Variable Moments
G. Work-Energy Applied to Plane Motion
A. Application of the Work-Energy Method.
Constant Forces
If a body is subjected to different sets of forces during different phases of its motion, the
resultant work summed up for all these phases may be equated directly to the total
change in kinetic energy. It is not necessary to compute the velocity at the end of any
phase so that it may be used as the initial velocity for the next phase. For example,
assume that the car of weight W lb in Fig. 14-3, starting with a velocity vo , reaches the
bottom of an incline 8 ft long with a velocity v1, and then rolls a distance s2 & ft along the
level to a position at which its velocity is .
A. Application of the Work-Energy Method.
Constant Forces
Writing the work- energy relation for the displacements &, and 8, we
have..

1𝑊 1𝑊
σ 𝑥 1 . 𝑠1 = 𝑣12 - 𝑣02
2 𝑔 2 𝑔
Equation A
1𝑊 1𝑊
σ 𝑥 2 . 𝑠2 = 𝑣22 - 𝑣12
2 𝑔 2 𝑔

Adding these two equations gives

1𝑊 1𝑊
σ 𝑥 1 . 𝑠1 + σ 𝑥 2 . 𝑠2 = ‫ ׬‬σ 𝑥 . ds = 𝑣22 - 𝑣02 Equation B
2 𝑔 2 𝑔
A. Application of the Work-Energy Method.
Constant Forces
Example :
The 300-lb block in Fig. 14-4a rests upon a level plane for which the oefficient of
kinetic friction is 0.20. Find the velocity of the block after it moves 80 ft, starting
from rest. If the 100-lb force is then removed, how much farther will it travel?
B. Resultant Work. Variable Forces
The general form of the work-energy equation has been shown

𝑠 𝑊
‫׬‬0 σ 𝑥 ⅆ𝑠 = ( v2 - 𝑣02 )
2𝑔
B. Resultant Work. Variable Forces
The force-displacement diagram for a spring in the figure is a straight line
determined from Hooke’s law and expressed by the equation

P = ks Equation A

in which k is known as the spring

modulus
B. Resultant Work. Variable Forces
This modulus or spring constant represents the force required to deform a given
spring through a unit distance.
𝑠
Noting that the resultant work, ‫׬‬0 σ 𝑥 ⅆ𝑠, is represented by the area under the
force-displacement diagram, we see' from Fig. 14-8 that the w.rk done in
deforming a spring from its free or unloaded length to an extension (or
compression) of s units is the area of the triangle OAB, or
𝑠 1 1
‫׬‬0 σ 𝑥 ⅆ𝑠 = (ks) (s) = 𝑘𝑠2 Equation b
2 2
𝑠
The work done may also be found by evaluating the integral ‫׬‬0 σ 𝑥 ⅆ𝑠, in which σ 𝑥
Is equal to ks.
This gives….
𝑠
𝑠 1 1
‫׬‬0 σ 𝑥 ⅆ𝑠 =න ks . ds = 𝑘𝑠2
2 2
0
B. Resultant Work. Variable Forces
The force-displacement diagram is especially useful in determining work required to
stretch a spring from an initial deformation s, to a larger deformation &. In this case, the
resultant work is the area of the trapezoid CDEF, which is equivalent to the average
force multiplied bye the change in deformation, or

RW = 𝑘𝑠1+𝑘𝑠2 (s – s )
2 1
2
B. Resultant Work. Variable Forces
Example :

At the instant shown in Fig. 14-10a, an external force has push the 500-lb block against the
spring, thereby compressing the spring 6 in. If the spring 1416. At the instant shown in Fig.
14-10a, an external force has pushed the constant is 100 lb per in., how far will the block
be projected along the level plane (for which fk is 0.20) when the external force is released?
What will be its maximum velocity?
C. Power. Efficiency
Power is the time rate at which work is done on a body. For example, if a train is being
pulled by a locomotive, the work done on the train be measured by the work done by the
drawbar pull. Expressed mathematically,

𝑊𝑜𝑟𝑘
Power =
𝑇𝑖𝑚𝑒
This gives the average power. If F is the net force doing the work, then work during any
instant is give by

σ 𝑥 . ds = F . ⅆs
and the power exerted at any instant is

𝑊𝑜𝑟𝑘 𝐹𝑑𝑠
Power = = = Fv
𝑇𝑖𝑚𝑒 𝑑𝑡
C. Power. Efficiency
Example :
A 1000-ton train is accelerated at a constant rate up a 2% grade. The train resistance is
constant at 10 lb per ton. The velocity increases from 20 mph to 40 mph in a distance of
2000 ft. Determine the maximum horsepower developed by the locomotive.
Solution
𝑊
σ𝑥. 𝑠 = ( v2 - 𝑣02 )
2𝑔

2 2 000 000
( P – 10 000 – 20 000 x ) x 200 = x (58.62 – 29.32)
100 64.4
P = 90 000 lb
The maximum power is developed when the train is moving at maximum speed
Therefore
F𝑣 90 000 𝑥 58.6
hp = hp = hp = 9600
550 550
D. Work-Energy Applied to Curvilinear
Translation
Many bodies are relatively so small compared with their paths of travel that they may be
treated as particles. This assumption applies to the curvilinear motion of all bodies
discussed in this article. The derivation of the work-energy equation for rectilinear
translation is also valid for curvilinear translation provided we assume that the X axis is
tangent to the path during each distance ds measured along the path. The resultant work is
𝑠 𝑊
then also ‫׬‬0 σ 𝑥 ⅆ𝑠 and the kinetic energy is 2𝑔
( v2 - 𝑣02 ) where v and v0, are the final and
initial velocities directed tangent to the path. The work done by gravity forces is best
computed as the weight multiplied by the change in the elevation (see Art.14-3) whereas
the work done by a spring may be found as in Art. 14-5.
D. Work-Energy Applied to Curvilinear
Translation
Example :
A 100-lb weight is swung in a vertical circle at the end of
a 4-ft cord. The topmost velocity of the weight is 12 ft
per sec. Find the tension in the cord at 120° past top
position. What minimum velocity at the bottom will keep
the weight in the circular path at the top?
E. Fundamental Work-Energy Equation
for Rotation
The fundamental equation of work-energy applied to rotation is ob- tained by the same
procedure as in translation. (See Art. 14-2.) We begin by eliminating a in the equations

Σ Μ = Ια (a)
and

a dθ = w . Dw (b)
and we obtain

Σ Μ . dθ = Iw dw (c)
E. Fundamental Work-Energy Equation
for Rotation
This is integrated between the limits or conditions that the angular dis- placement varies
from zero to a final value of θ while the angular velocity changes from an initial value of wo,
to a final value of w.
We thus obtain

𝜃 𝑤
‫׬‬0 Σ 𝑀 . ⅆθ = I ‫𝑤 𝑤׬‬ ⅆ𝑤
𝑜
Or

𝜃 1 2 1 2 1
‫׬‬0 Σ 𝑀 . ⅆθ = Iw - I𝑤0 = I (w2 - 𝑤02 )
2 2 2

This is the fundamental work-energy equation for rotation

E. Fundamental Work-Energy Equation
for Rotation
As in translation, no acceleration is involved; hence the resultant work on the body
(however computed) during any motion is always equivalent to the resultant change in
kinetic energy. Also, as in translation, work is considered positive in the direction of motion;
i.e., moments in the direction of motion do positive work on the body, and vice versa.
E. Fundamental Work-Energy Equation
for Rotation
The expression for the kinetic energy of a rotating body can also be developed from the
fact that the total kinetic energy is the summation of the kinetic energies of all particles of
the body. Thus in Fig. 14-13, if A is a typical particle, its kinetic energy is

1 𝑑𝑊 2 1 𝑑𝑊 2 2
KE = 𝑉𝐴 = r w
2 𝑔 2 𝑔
For the Entire body

Total KE = 1 𝑑𝑊 1 2 𝑑𝑊
න 𝑔 r w w
2 2 = න𝑟2
2 2 𝑔

Or, as before

1
KE = 𝐼 𝑤2
2
F. Work-Energy Method with Constant
Moments
In the following illustrative problems, the forces causing rotation are constant. Since the
moment arms of these forces are also constant, rotation is due to constant moment.
directly to a system of connected bodies without considering any of the.
As in the case of translation, the work-energy equation may be applied internal forces in the
connecting elements. Then after the relation between force, displacement, and velocity for
the original system has been deter- mined, we may isolate any part of the system and
reapply the work-energy equation to that part.
F. Work-Energy Method with Constant
Moments
Determine the distance that body D in Fig. 14-14 must move in order to reach a velocity of
24 it per sec starting from rest: What tension is acting in the cord joining the step pulleys B
and C during this movement?
G. Work-Energy Method with Variable
Moments
Although the forces causing rotation may be constant in the following illustrative problems,
their moment arms are not; this causes rotation under the action of a variable moment.
Angular velocities in such cases are best obtained by the work-energy method which
eliminates consideration of variable angular acceleration.
The work done by gravity forces, will be the product of the weight W by
its change in elevation h as discussed in Art. 14-3. It is worth while to
check this statement by considering Fig. 14-17 in which the body rotates
through θradians from the horizontal. Although not so convenient, the
work done by W can be found by applying ‫׬‬0 Σ 𝑀 . d
𝜃
θ which gives

‫׬‬0 Σ 𝑀 . d θ = ‫׬‬0 W 𝑟Ԧ cos θ dθ = W 𝑟 [ sin θ ]𝜃0 = W 𝑟Ԧ sin θ

𝜃 𝜃
Ԧ
and as 𝑟Ԧ sin θ = h, we finally obtain Wh as stated before.
G. Work-Energy Method with Variable
Moments
1455. A uniform rod 6 ft long weighing 96.6 lb carries a 32.2-lb weight at one end. As shown
in Fig. 14-18, it starts from rest in a horizontal position and rotates through a vertical are of
60°. At this instant determine the bearing reaction at A
H. Work-Energy Applied to Plane Motion
• We have shown In Art. 13-16 that a plane motion is equivalent to a combination of pure
rotation about its centroidal axis and a translation of its center gravity
• For the translational component of plane motion, tak the X axis tangent to the path of the
gravity center during each displacement ⅆ 𝑠Ԧ along the path, we have
𝑠 𝑊
RW = ‫׬‬0 σ 𝑥 ⅆ𝑠 = ( v2 - 𝑣02 ) Equation A
2𝑔

And for the rotational moment

𝜃 1
RW = ‫׬‬0 Σ 𝑀 . ⅆθ = I (w2 - 𝑤02 ) Equation B
2
Equating the sum of 2 equation

𝑊 1
RW = ( v2 - 𝑣02 ) + I (w2 - 𝑤02 )
2𝑔 2
H. Work-Energy Applied to Plane Motion
If the instant center of plane motion can be located, it is usually mor convenient to express the kinetic
energy in terms of an instantaneou rotation about the instant center. Thus expressing the kinetic energy any
instant as

1𝑊 2 1
KE = 𝑣Ԧ + I w2
2𝑔 2
We may replace 𝑣Ԧ by 𝑟𝑤
Ԧ is the distance from the instant center to center of gravity. Doing this
1𝑊 1 1 𝑊
gives KE = 𝑟Ԧ 2 𝑤 2 + I w2 = w2 (I+ 𝑟Ԧ 2 )
2 𝑔 2 2 𝑔
𝑊
since, by the transfer formula I + 𝑟Ԧ 2 = Ic which is the mass moment of inertia about the instant
𝑔
center C.
H. Work-Energy Applied to Plane Motion
Except for the free-rolling wheels, the position of the instant center relative to the body usually
changes, so that the change in kinetic energy must be expressed as
1 1
Δ𝐾𝐸 = 2 𝐼𝐶 𝑤 2 − 𝐼𝐶0 𝑤02
2
in which Ic and Ico representively the moments of inertia about the final and initial positions of the
instant center C.

final andThis last expression suggests a special but convenient form of the work- energy equation for
homogeneous free-rolling wheels in which Ic is constant. Considering the plane motion as equivalent
to rotation about the instant center, we may write
1
σ 𝑀𝑐 = 𝐼 𝑤 2 − 𝑤02
2 𝑐
H. Work-Energy Applied to Plane Motion
The system in Fig. 14-22 has the values shown.. Assuming disk D to roll without slipping, determine
the linear velocity of its center after body B has moved 12 ft, starting from rest. Also determine the
friction force acting on the disk D to prevent slipping.