Work Energy & Power - Lecture notes 7 PDF

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Detailed lecture on Work, Power & Energy and it's theoram...

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WORK, ENERGY & POWER

Terms Energy The ability to do work. Work A force applied over a distance. For formulae, see work done by a constant force parallel to displacement and work done by any constant force, and work done by a position-dependent force. Joule The units of work, equivalent to a Newton-meter. Also, units of energy. Kinetic Energy The energy of motion. Power Work done per unit time. For formulas, see Formula for average power, Definition of instantaneous power, and formula for instantaneous power. Watt Unit of power; equal to joule/second. Conservative force: Any force which conserves mechanical energy, as opposed to a nonconservative force. See statement of conservation of mechanical energy.

Conservative System: A system in which energy is conserved.

Nonconservative Force: Any force which does not conserve mechanical energy, as opposed to a conservative force.

Path independence: Property of conservative forces which states that the work done on any path between two given points is the same.

Potential energy : The energy of configuration of a conservative system. For formulae, see Definition of potential energy, gravitational potential energy, and Definition of potential energy given a position-dependent force.

Total mechanical energy: The sum of the kinetic and potential energy of a conservative system. See definition of total mechanical energy.

Conservative force: Any force which conserves mechanical energy, as opposed to a nonconservative force. See statement of conservation of mechanical energy.

Conservative System: A system in which energy is conserved.

Energy: The ability to do work.

Non-conservative Force: Any force which does not conserve mechanical energy, as opposed to a conservative force.

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Formulae Work done by a constant force parallel to displacement

W = Fx

Work done by any constant force

W = Fx cosθ

Work-Energy Theorem

W = ΔK

Formula for average power

Definition of instantaneous power

=

P=

Formula for instantaneous power

P = Fv cosθ

Work done by a position-dependent force

W=

Definition of potential energy.

F(x)dx force.

ΔU = - W

Gravitational potential energy.

UG = mgh

Statement of conservation of mechanical energy.

Δ(U+K) = 0

Definition of total mechanical energy.

U+K=E

Definition of potential energy given a position-dependent force.

ΔU = -

Work done by a constant force parallel to displacement

W = Fx

Work done by any constant force

W = Fx cosθ

Work-Energy Theorem

W = ΔK

Formula for average power

F(x)dx

=

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Definition of instantaneous power

P=

Formula for instantaneous power

P = Fv cosθ

Work done by a position-dependent force

W=

Definition of potential energy.

ΔU = - W

Gravitational potential energy.

UG = mgh

Statement of conservation of mechanical energy.

Δ(U+K) = 0

Definition of total mechanical energy.

U+K=E

Definition of potential energy given a position-dependent force.

ΔU = -

F(x)dx force.

F(x)dx

Introduction and Summary Newton's Laws, and dynamics as a whole, provide us with fundamental axioms for the study of classical mechanics. Once these foundations are laid, we can derive new concepts from the axioms, furthering our understanding of mechanics and allowing us to extend our study to new and more complex physical situations. Perhaps the most significant concept derived in dynamics is that of work. The understanding of work greatly simplifies many physical situations. Work, in a sense, introduces a dynamic understanding to mechanics. It allows us to evaluate forces over distance and time, to give us a broader understanding not just of the forces acting on a given object, but about what happens to that object over the course of a given journey. In addition, the concept of work makes our complicated kinematics equations virtually obsolete. It makes calculations easier and allows us to extend our study to other realms. We will begin by defining work, both mathematically and conceptually. Once we have an understanding of work, we can apply it to a new concept, energy, the measure of change within a system, and establish the Work-Energy Theorem. We will also look at power, a practical concept that is derived from work. Finally, as we begin to explore more complex situations, we will examine work from a calculus standpoint, and examine variable forces. A good understanding of work is essential for further studies in physics. Work is not only a gateway to more advanced mechanics concepts, it is a concept used in all areas of physics.

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Definition of Work Work, though easily defined mathematically, takes some explanation to grasp conceptually. In order to build an understanding of the concept, we begin with the simplest situation, then generalize to come up with the common formula.

The Simple Case Consider a particle moving in a straight line that is acted on by a constant force in the same direction as the motion of the particle. In this very simple case, the work is defined as the product of the force and the displacement of the particle. Unlike a situation in which you hold something in place, exerting a normal force, the crucial aspect to the concept of work is that it defines a constant force applied over a distance. If a force F acts on a particle over a distance x, then the work done is simply: W = Fx Since w increases as x increases, given a constant force, the greater the distance during which that force acts on the particle, the more work is done. We can also see from this equation that work is a scalar quantity, rather than a vector one. Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account.

What are the units of work? The work done by moving a 1 kg body a distance of 1 m is defined as a Joule. A joule, in terms of fundamental units, is easily calculated:

W = Fx =

(m) =

The joule is a multipurpose unit. It serves not only as a unit of work, but also of energy. Also, the joule is used beyond the realm of physics, in chemistry, or any other subject dealing with energy.

In dynamics we were able to define a force conceptually as a push or a pull. Such a concise definition is difficult to generate when dealing with work. To give a vague idea, we can describe work as a force applied over a distance. If a force is to do work, it must act on a particle while it moves; it cannot just cause it to move. For instance, when you kick a soccer ball, you do no work on the ball. Though you produce a great deal of motion, you have only instantaneous contact with the ball, and can do no work. On the other hand, if I pick the ball up and run with it, I do work on the ball: I am exerting a force over a certain distance. In technical jargon, the "point of application" of the force must move in order to do work. Now, with a conceptual understanding of work, we can move on to define it generally.

The General Case In the last section we came up with a definition of work given that the force acted in the same direction as the displacement of the particle. How do we calculate work if this is not the case? We simply resolve the force into components parallel and perpendicular to the direction of displacement of the particle (see Vectors, Component Method). Only the force parallel to the displacement does work on the particle. Thus, if a force is applied at an angle θ to the displacement of the particle, the resulting work is defined by: W = (F cosθ) x

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This new equation has similar form to the old equation but provides a more complete description. If θ = 0, then cosθ = 1 and we have our first equation. Also, this equation ensures that it does not take into account any forces acting on a moving particle that do not do any work. Consider the normal force acting on a ball rolling across a horizontal floor. The normal force is perpendicular to the motion, implying that θ = 90 and cosθ = 0. Thus, there is no work done on the ball by the normal force. In this sense, work can be seen as produced by any force that aids or hinders the motion of the particle. Stationary forces and forces perpendicular to the motion do not cause work. To see an example of work in a simple system, let us consider the work done by a gravitational force on a falling object. The gravitational force is simply mg, and let us denote the distance of the fall by h. Clearly if the object simply falls straight down, the work done is given by W = mgh. But what if the object falls at an angle θ from vertical, as seen below?

Figure %: An object falling at an angle If the object falls the same height, then the distance travelled is given by x =

W = Fx cosθ = (mg)(

. The work, then, is given by:

)(cosθ) = mgh

As long as the object falls h distance, the work done on the object falling at an angle is the same as if the object were falling straight down. This fact, special to gravity and other forces, is significant in the study of energy, but for now suffices to demonstrate how to calculate work.

Work is commonly misunderstood because of its common definition. Most people think that it takes a lot of work to hold a 100-pound weight in the air. The weight is not moving, though, so in the sense of physics no work is done. It is important to realize how our definition differs from a common one and stick to the physical understanding of work. From this definition of work, we will be able to bring in a concept of energy, and greatly simplify many aspects of classical mechanics.

Problems Problem: A 10 kg object experiences a horizontal force which causes it to accelerate at 5 m/s2, moving it a distance of 20 m, horizontally. How much work is done by the force? The magnitude of the force is given by F = ma = (10)(5) = 50 N. It acts over a distance of 20 m, in the same direction as the displacement of the object, implying that the total work done by the force is given by W = Fx = (50) (20) = 1000 Joules.

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Problem: A ball is connected to a rope and swung around in uniform circular motion. The tension in the rope is measured at 10 N and the radius of the circle is 1 m. How much work is done in one revolution around the circle? Recall from our study of uniform circular motion that centripetal force is always directed radially, or toward the centre of the circle. Also, of course, the displacement at any given time is always tangential, or directed tangent to the circle:

Work in Uniform Circular Motion Clearly the force and the displacement will be perpendicular at all times. Thus, the cosine of the angle between them is 0. Since W = Fx cosθ, no work is done on the ball.

Problem: A crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. The tension in the rope is 50 N. How much work is done in moving the crate 10 meters? In this problem a force is exerted which is not parallel to the displacement of the crate. Thus, we use the equation W = Fx cosθ. Thus W = Fx cosθ = (50) (10) (cos 30) = 433 J

Problem: A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight? The crate, and thus the point of application of the force, does not move. Thus, though a force is applied, no work is done on the system.

Problem: A 5 kg block is moved up a 30-degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is .25. How much work is done by the 50 N force in moving the block a distance of 10 meters? What is the total work done on the block over the same distance? Finding the work done by the 50 N force is quite simple. Since it is applied parallel to the incline, the work done is simply W = Fx = (50) (10) = 500 J.

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Finding the total work done on the block is more complex. The first step is to find the net force acting upon the block. To do so we draw a free body diagram:

Work on an Incline Because of its weight, mg, the block experiences a force down the incline of magnitude mg sin 30 = (5) (9.8) (.5) = 24.5 N. In addition, a frictional force is felt opposing the motion, and thus down the incline. Its magnitude is given by Fk = μFN = (.25) (mg cos 30) = 10.6 N. In addition, the normal force and the component of the gravitational force that is perpendicular to the incline cancel exactly. Thus, the net force acting on the block is: 50 N -24.5 N -10.6 N = 14.9 N, directed up the incline. It is this net force that exerts a net work on the block. Thus, the work done on the block is W = Fx = (14.9) (10) = 149 J.

The Work-Energy Theorem Now that we have a definition of work, we can apply the concept to kinematics. Just as force was related to acceleration through F = ma, so is work related to velocity through the Work-Energy Theorem.

Derivation of the Work-Energy Theorem It would be easy to simply state the theorem mathematically. However, an examination of how the theorem was generated gives us a greater understanding of the concepts underlying the equation. Because a complete derivation requires calculus, we shall derive the theorem in the one-dimensional case with a constant force. Consider a particle acted upon by a force as it moves from xo to xf. Its velocity also increases from vo to vf. The net work on the particle is given by:

Wnet = Fnet(xf - xo) Using Newton's Second Law we can substitute for F:

Wnet = ma (xf - xo) Given uniform acceleration, vf2 - vI2 = 2a (xf - xo). Substituting for a (xf - xo) into our work equation, we find that:

Wnet =

mvf2 -

mvo2

This equation is one form of the work-energy equation and it gives us a direct relation between the net work done on a particle and that particle's velocity. Given an initial velocity and the amount of work done on a particle, we can calculate the final velocity. This is important for calculations within kinematics, but is even more important for the study of energy, which we shall see below. SYED FATEHYAB MEHDI

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Kinetic Energy and the Work-Energy Theorem As is evident by the title of the theorem we are deriving, our ultimate goal is to relate work and energy. This makes sense as both have the same units, and the application of a force over a distance can be seen as the use of energy to produce work. To complete the theorem, we define kinetic energy as the energy of motion of a particle. Taking into consideration the equation derived just previously, we define the kinetic energy numerically as:

mv2

K.E = Thus, we can substitute K in our work energy theorem:

Wnet =

mvf2 -

mvI2 = Kf - Ko

Implying that

Wnet = ΔK This is our complete Work-Energy theorem. It is powerfully simple and gives us a direct relation between net work and kinetic energy. Stated verbally, the equations say that net work done by forces on a particle causes a change in the kinetic energy of the particle.

Though the full applicability of the Work-Energy theorem cannot be seen until we study the conservation of energy, we can use the theorem now to calculate the velocity of a particle given a known force at any position. This capability is useful, since it relates our derived concept of work back to simple kinematics. A further study of the concept of energy, however, will yield far greater uses for this important equation.

Proble Problems ms Problem: What is the kinetic energy of a 2 kg ball that travels a distance of 50 meters in 5 seconds?

The velocity of the ball is easily calculable: v = ball, we can calculate kinetic energy:

K=

mv2 =

= 10 m/s. With values for the mass and velocity of the

(2 kg) (10 m/s)2 = 100 J

Problem: In a sense we all have kinetic energy, even when we are standing still. The earth, with a radius of 6.37×106 meters, rotates about its axis once a day. Ignoring the earth's rotation about the sun, what is the kinetic energy of a 50 kg man standing on the surface of the earth? To find the velocity of the man we must find how far he travels over a given time period. In one day, or 86400 seconds, the man travels the circumference of the earth, or 2Πr meters. Thus, the velocity of the man is v =

=

kinetic energy. K =

= 463 m/s. Again, since we know the velocity and mass of the man we can calculate mv2 =

(50 kg) (463 m/s)2 = 5.36×106 Joules.

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Problem: A ball is dropped from a height of 10 m. What is its velocity when it hits the ground? The ball is acted upon by a constant gravitational force, mg. The work done during its total trip, then, is simply mgh. By the Work-Energy theorem, this causes a change in kinetic energy. Since the ball initially has no velocity, we can find the final velocity by the equation: W = ΔK

mgh =

mv2

Solving for v, v=

=

= 14 m/s

The final velocity of the ball is 14 m/s. We found this by one simple calculation, avoiding the cumbersome kinematic equations. This is an excellent demonstration of the advantages of working with the concepts of work and energy, as opposed to simple kinematics.

Problem: A ball is thrown vertically with a velocity of 25 m/s. How high does it go? What is its velocity when it reaches a height of 25 m? The ball reaches its maximum height when its velocity is reduced to zero. This change in velocity is caused by the work done by gravitational force. We know the change in velocity, and hence the change in kinetic energy of the ball, and can calculate its maximum height from this: W = ΔK

mgh =

mvf2 -

mvo2

But vf = 0, and the masses cancel, so

h=

=

= 31.9 m

When the ball is at a height of 25 meters, the gravitational force has done an amount of work on the ball equal to W = - mgh = - 25 mg. This work causes a change in velocity of the particle. We now use the WorkEnergy Theorem, and solve for the final velocity:

- mgh =

mvf2 -

mvo2

Again, the masses cancel:

vf2 =

vo2 - gh

Thus Vf =

=

= 11.6 m/s .. SYED FATEHYAB MEHDI

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Problem: A ball with enough speed can complete a vertical loop. With what speed must the ball enter the loop to complete a 2 m loop? (Keep in mind that the velocity of the ball is not constant throughout the loop). At the top of the loop, the ball must have enough velocity such that the centripetal force provided by its weight keeps the ball in circular motion. In other words:

FG = Fc thus mg = Solving for v, v=

=

= 4.4 m/s

This value for the velocity gives us the minimum velocity at the top of the loop. But we are asked for the minimum velocity at the bottom of the loop. How do we find this? You guessed it: Work-Energy Theorem.