Workand
Energy
• A force F does work on a particle only when the particle undergoes adisplacement in the direction of the force.
The Work of a Force
Consider the force acting on the particle
• If the particle moves along the path from position r to new position r’ (= r+dr),displacement dr = r’ – r
• ds is the magnitude of dr
• Resultant interpreted in two ways1. Product of F and the component of displacement
in the direction of the force ds cos θ2. Product of ds and component of force in the
direction of the displacement F cos θ
• θ is the angle between tails of dr and F
• Work dU done by F is a scalar quantity
dU = F ds cos θdU = F·dr
The Work of a Force
• If 0° < θ < 90°, the work is positive
• If 90° < θ < 180°, the work is negative
• dU = 0 if θ = 90 °, or displacement = 0
Basic unit for work in SI units is joule (J)• 1 joule of work is done when a force of 1 Newton
moves 1 meter along its line of action1J = 1N.m
The Work of a Force
Work of a Variable Force.
• If the particle undergoes a finite displacement along its path from r1 to r2 or s1 to s2, the work is determined by integration.
• If F is expressed as a function of position, F = F(s),
2
1
2
1cos.21
s
s
r
rdsFdU rF
The Work of a Force
• The work can be interpreted as the area under the curve from position s1 to position s2
2
1
2
1cos.21
s
s
r
rdsFdU rF
The Work of a Force
Work of a Constant Force Moving Along a Straight Line.
)(cos
cos
12
212
1
ssF
dsFU
c
s
sc
The work of Fcrepresents the area of the rectangle
The Work of a Force
Work of a Weight.
Force W = -Wj
Displacement
dr = dxi +dyj + dzk.
yWU
yyWWdy
kdzjdyidxjW
rdFU
y
y
r
r
21
12
21
)(
)).((
.
2
1
2
1
Independent of the path
The Work of a Force
Work of a Spring Force.
Force on a linear elastic spring
Fs = ks.
k: the spring stiffness
s: distance from its unstretched position • If the spring is elongated or compressed from a
position s1 to s2, the work done on spring by Fs is always positive, since force and displacement are in the same direction.
The Work of a Force
)(21
21
21
21
22
21
22
212
1
2
1
ssk
ksks
dsksdsFUs
s
s
s s
• This equation represents the trapezoidal area under the line Fs = ks
The Work of a Force
Example 1The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m.
Free Body Diagram
•Horizontal Force P.
•Spring Force Fs.
•Weight W.
•Normal Force NB.
s
Example 1
Example 1
ssExample 1
• Principle of work and energy for the particle,
2211
1221
2221
or 21
21
TUT
TTmvmvU
• U1-2: the sum of work done by all the forces acting on the particle as the particle moves from point 1 to point 2
• T1: the particle’s initial kinetic energy.• T2: the particle’s final kinetic energy.
Principle of Work and Energy
Principle of Work and Energy • For the particle P moving
in the tangential direction, ∑Ft = mat
• Using ads = vdv and integrating over ds
21
22 2
121
2
1
2
1
mvmv
dvmvdsFv
v
s
s t
• Kinetic Energy of a particle 2
21 mvT
POWER AND EFFICIENCYEngines and motors are often rated in terms of their power output. The power requirements of the motor lifting this elevator depend on the vertical force F that acts on the elevator, causing it to move upwards.
Given the desired lift velocity for the elevator, how can we determine the power requirement of the motor?
APPLICATIONSThe speed at which a vehicle can climb a hill depends in part on the power output of the engine and the angle of inclination of the hill.
For a given angle, how can we determine the speed of this jeep, knowing the power transmitted by the engine to the wheels?
POWER
Thus, power is a scalar defined as the product of the force and velocity components acting in the same direction.
Since the work can be expressed as dU = F • dr, the power can be written
P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
If a machine or engine performs a certain amount of work, dU, within a given time interval, dt, the power generated can be calculated as
P = dU/dt
Power is defined as the amount of work performed per unit of time.
POWERUsing scalar notation, power can be written
P = F • v = F v cos where is the angle between the force and velocity vectors.
So if the velocity of a body acted on by a force F is known, the power can be determined by calculating the dot product or by multiplying force and velocity components.
The unit of power in the SI system is the watt(W) where
1 W = 1 J/s = 1 (N ·m)/s .
EFFICIENCY
If energy input and removal occur at the same time, efficiency may also be expressed in terms of the ratio of output energy to input energy or
= (energy output)/(energy input)
Machines will always have frictional forces. Since frictional forces dissipate energy, additional power will be required to overcome these forces. Consequently, the efficiency of a machine is always less than 1.
The mechanical efficiency of a machine is the ratio of the useful power produced (output power) to the power supplied to the machine (input power) or
= (power output)/(power input)
PROCEDURE FOR ANALYSISWork (Free-Body Diagram)• Establish the initial coordinate system and draw a FBD of the particle to account for all the forces that do work on the particle as it moves along its path
Principle of Work and Energy• Apply the principle of work and energy
• The kinetic energy at the initial and final points is always positive since it involves the speed squared
2211 TUT
• A force does work when it moves through a displacement in the direction of the force
• Work is always positive when the force component is in the same direction as its displacement, otherwise, it is negative
• Forces that are functions of displacement must be integrated to obtain the work
• Graphically, the work is equal to the area under the force-displacement curve
PROCEDURE FOR ANALYSIS
• The work of a weight is the product of the weight magnitude and the vertical displacement
• It is positive when the weight moves downwards• The work of the spring is in the form of
where k is the spring stiffness and s is the stretch or compression of the spring
221 ksUs
PROCEDURE FOR ANALYSIS
The 17.5-kN automobile is traveling down the 10°inclined road at a speed of 6 m/s. If the driver jams on the brakes, causing his wheels to lock, determine the distance s his tires skid on the road. The coefficient of the kinetic friction between the wheels and the road is μk = 0.5
Example 2
Example 2
Example 2
The platform P is tied down so that the 0.4-m long cords keep a 1-m long springcompressed 0.6-m when nothing is on the platform (Fig. a). If a 2-kg block is placed on the platform and released from rest after the platform is pushed down 0.1-m (Fig. b), determine the max height h the block rises in the air, measure from the ground.
Example 3
Work (Free-Body Diagram).• The initial and final velocities are zero. • The weight does negative work and the spring force
does positive work.
Example 3
Example 3
- From Fig below, we can write the weight of an object at a distance r from the center of the earth in terms of polar coordinates.
rE
rmgR eF 2
2
Work Done by Particular Forces
- Using the expression for the velocity in terms of polar coordinates:
• So the dot product of F and dr is:
eeeer drdrdtdtdr
dtdrd rr
drr
mgR
drdrr
mgRdF
E
rrE
2
2
2
2eeer
Work Done by Particular Forces
Thus the work reduces to an integral w.r.t. r:
12
2
r
r 2
2r
r12
11
2
1
2
1
rrmgR
drr
mgRdrFU
E
E
Work Done by Particular Forces
• Pumped-storage hydropower plant
Conservative Forces and Potential Energy
Conservative Force.
• It is defined by the work done in moving a particle from one point to another that is independent of the path followed by the particle.
Potential Energy. (capacity of an object to do work)
• It is the measure of the amount of work a conservative force will do when it moves from a given position to the datum.
Conservative Forces and Potential Energy
datum
Gravitational Potential Energy.
• A datum is arbitrarily selected as a reference level.
• If y is positive upward, gravitational potential energy of the particle of weight W is
Vg = +Wy
+h
-h
Vg=+Wh
Vg= - Wh
Conservative Forces and Potential Energy
• At the datum Vg = 0.
• Above the datum Vg > 0.
• below the datum Vg < 0
Elastic Potential Energy
When an elastic spring with stiffness k is elongated or compressed a distance s from its unstretched position,
2
0
20
21
21d
ks
kxxkxVss
e
Ve is always positive
Conservative Forces and Potential Energy
Potential Function.
• If a particle is subjected to both gravitational and elastic forces, the particle’s potential energy can be expressed as a potential function
221 ksWy
VVV eg
Conservative Forces and Potential Energy
Conservative of Energy• Work done by conservative forces written in
terms of the difference in their potential energies
21.21 VVU cons
• The principle of work and energy can be written as
22.2111 )( VTUVT noncons
• Work done by non-conservative forces
.21 )( nonconsU
• If only conservative forces are applied, we have the law of the conservation of mechanical energy
2211 VTVT
• It is used to solve problem involving velocity, displacement and conservative force systems.
Conservative of Energy
Potential Energy.
• Draw two diagrams showing the particle located at its initial and final points along the path
• If the particle is subjected to a vertical displacement, establish the fixed horizontal datum.
PROCEDURE FOR ANALYSIS
• Determine the elevation y of the particle from the datum and the extension or compression s of any connecting springs
• Gravitational potential energy Vg = Wy
• Elastic potential energy 221 ksVe
Conservation of Energy
• Apply the equation
2211 VTVT
• When determining the kinetic energy, the particle’s speed v must always be measured from an inertial reference frame.
221 mvT
PROCEDURE FOR ANALYSIS
Example 4
The gantry structure is used to test the response of an airplane during a crash. The plane of mass 8-Mg is hoisted back until θ = 60°, and then pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before crashing into the ground, θ = 15°. Also, what is the maximum tension developed in the supporting cable during the motion?
Potential Energy.
For convenience, the datum has been established at the top of the gantry.
Example 4
8-Mg
60°
15°
Example 4
Example 4
The ram R has a mass of 100-kg and is released from rest 0.75-m from the top of a spring, A, that has a stiffnesskA = 12 kN/m. If a second spring B, having a stiffness kB= 15 kN/m is “nested” in A, determine the max displacement of A needed to stop the downward motion of the ram.
Example 5
Potential Energy.
• The datum is located through the center of gravity of the ram at its initial position.
• Assume that the ram compresses both springs at the instant it comes to rest (v2 = 0).
• A is compressed by sA
• B by sB = sA – 0.1 m
Example 5
Example 5
Example 5