1. The Work of a Force
•A force F does work on a particle only when the particle undergoes a displacement in
the direction of the force.
•Consider the force acting on the particle
•If the particle moves along the path s from position r to new position r’, displacement
dr = r’ – r
•Magnitude of dr is represented by ds, differential segment along the path
•If the angle between tails of dr and F is θ, work dU done by F is a scalar quantity
dU = F ds cos θ
dU = F·dr
•Resultant interpreted in two ways
 Product of F and the component of displacement in the direction of the force
ds cos θ
 Product of ds and component of force in the direction of the displacement F cos θ
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2. •If 0° < θ < 90°, the force component and the displacement has the same sense so that
the work is positive
•If 90° < θ < 180°, the force component and the displacement has the opposite sense so
that the work is negative
• dU = 0 if the force is perpendicular to the displacement since cos 90° = 0 or if the force
is applied at a fixed point where displacement = 0
•Basic unit for work in SI units is joule (J)
•This unit combines the units for force and displacement
•1 joule of work is done when a force of 1 newton moves 1 meter along its line of action
1J = 1N.m
•Moment of a force has this same combination of units, however, the concepts of
moment and work are in no way related
•A moment is s vector quantity, whereas work is a scalar
Work of a Variable Force.
•If the particle undergoes a finite displace along its path from r1 to r2 or s1 to s2, the
determined by integration.
•If F is expressed as a function of positio
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3. •If the working component of the force, F cos θ, is plotted versus s, the integral in this
equation can be interpreted as the area under the curve from position s1 to position s2
Work of a Constant Force Moving Along a Straight Line.
•If the force Fc has a constant magnitude and acts at a constant angle θ from its straight
line path, then the components of Fc in the direction of displacement is Fc cos θ
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4. •The work done by Fc when the particle is displaced from s1 to s2 is determined
The work of Fc represents the area of the rectangle
Work of a Weight.
•Consider a particle which moves up along the path s from s1 to position s2.
•At an intermediate point, the displacement dr = dxi +dyj + dzk. Since W = -Wj
•Work done is equal to the magnitude of the particle’s weight times its vertical
displacement.
•If W is downward and ∆y is upward, work is negative
•If the particle is displaced downward (-∆y), the work of the weight is positive.
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5. Work of a Spring Force.
•The magnitude of force developed in a linear elastic spring when the spring is
displaced a distance s from its unstretched position is Fs = ks.
•If the spring is elongated or compressed from a position s1 to s2, the W.D on spring by
Fs is positive, since force and displacement are in the same direction.
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6. •If a particle is attached to a spring, then the force Fs exerted on the particle is opposite
to that exerted on the spring.
•The force will do negative work on the particle when the particle is moving so as to
further elongate (or compress) the spring.
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7. Example
The 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.
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8. Principle of Work and Energy
•Consider a particle P, which a the instant considered located on the path as measured
from an inertial coordinate system
•For the particle in the tangential direction, ∑Ft = mat
•For principle of work and energy for the particle,
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9. •Term on the LHS is the sum of work done by all the forces acting on the particle as the
particle moves from point 1 to point 2
•Term on the RHS defines the particle’s final and initial kinetic energy
•Both terms are always positive scalars
•The particle’s initial kinetic energy plus the work done by all the forces acting on the
particle as it moves from initial to its final position is equal to the particle’s final kinetic
energy
•For example, if a particle’s initial speed is known and the work of all the forces acting
on the particle can be determined, the above eqn provides a direct means of obtaining
the final speed v2 of the particle after it undergoes a specified displacement
PROCEDURE FOR ANALYSIS
Work (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
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10. •The kinetic energy at the initial and final points is always positive since it involves the
speed squared
•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
•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
•Principle of work and energy can be extended to include a system of n particles
isolated within an enclose region of space
•An arbitrary ith particle, having a mass mi, is subjected to a resultant force Fi and a
resultant internal force fi , which each of the other particles exerts on the ith particle
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11. • Principle of work and energy for the ith particle
• Since both work and force are scalars, the results may be added together
algebraically
• We can write this equation symbolically
• Equation states that the system’s initial energy plus the work done by all the
external and internal forces acting on the particles of the system is equal to the
system’s final kinetic energy
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12. • Internal forces on adjacent articles will occur in equal but opposite collinear pairs,
the total work done by each of these forces will not cancel out since paths over
which corresponding particles travel will be different
Two important exception:
1.if the particles are contained within the boundary of a translating rigid body, the
internal forces will undergo the same displacement and therefore the internal work will
be zero
2.Adjacent particles exert equal but opposite internal forces that have components
which undergo the same displacement and therefore the work of these forces cancels
Work of Friction Caused by Sliding.
•Consider a case when a body is sliding over the surface of another body in the
presence of friction
•Consider a block translating a distance s over a rough surface
• If the applied force P just balances the resultant frictional force μkN then due to
equilibrium a constant velocity v is maintained
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13. • Sliding motion will generate heat, a form of energy which seems not to be accounted
for in the work energy equation
• Model the block so that the surfaces of contact are deformable (nonrigid)
• Rough portions at the bottom of the block act as teeth and when the block slides
these teeth deform slightly and either break off or vibrate due to interlocking effects
and pull away from teeth at the contracting surface
• As a result, frictional forces that act on the block at these points are displaced
slightly, due to localized deformations, and then they are replaced by other frictional
forces as other points of contact are made
• At any instant, the resultant F of these frictional forces remain essentially constant,
μkN, however, due to many localized deformations, the actual displacement s’ of
μkN is not the same displacement s as the applied force P
• s’ < s and the external work done by the resultant force will be μkNs’ and not μkNs
the remaining work μkN(s –s’) manifests itself as the increase in internal energy
which in fact, causes the block temperature to increase
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14. Example
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 how far s his tires
skid on the road. The coefficient of the kinetic friction between the wheels and the road
is μk = 0.5
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15. Example
For a short time the crane lifts the 2.50-Mg beam with a force of F = (28 + 3s2) kN.
Determine the speed of the beam when it has risen s = 3 m. How much time does it
take to attain this height starting from rest.
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16. Example
The platform P is tied down so that the 0.4-m long cords keep a 1-m long spring
compressed 0.6-m when nothing is on the platform. If a 2-kg platform is placed on the
platform and released from rest after the platform is pushed down 0.1-m, determine the
max height h the block rises in the air, measure from the ground.
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