Chapter 5 Work and Energy

Transcription

1 Chapter 5 Work and Energy Work and Kinetic Energy Work W in 1D Motion: by a Constant orce by a Varying orce Kinetic Energy, KE: the Work-Energy Theorem Mechanical Energy E and Its Conservation Potential Energy Gravitational Potential Energy, PE g Elastic orces, g. Elastic Potential Energy, PE e Power P

2 Why an alternative treatment based on Energy and Momentum? In this lecture set, we ll refer to the same physical situations as in the previously, but from a different perspective: using energy Energy and momentum are physical quantities that obey conservation laws such that they can be used to describe how the motion is transmitted between interacting objects while overall they stay constant Conservation laws are still valid even when classical mechanics fails Ex: a) When an object moves along a rough surface its energy is transmitted to the ground so, if there is no other source to replenish its energy, the object will eventually stop: the total energy in the system object-surface is conserved. b) A freely falling object interacts with the Earth which transfers energy to the object such that the object will move faster and faster: the total energy in the system object-earth is conserved. In the previous lectures, we defined conservation of momentum. In this lecture we ll study conservation of energy: If a system is isolated, its total energy that is, the sum of all its energies present in various forms remains the same, albeit the form of energies may change In mechanics we are particularly interested in the energy associated with motion Kinetic Energy and how it changes via the Work done on the system by forces.

3 Work Done by a constant force in a 1D motion A force acting on an object may intermediate an exchange of energy measured by the work done by the force: a scalar quantity with dimension of energy The work done by a force depends on the force, on the displacement of the object, and on whether the force contributes to the respective displacement The motion of an object is characterized by a unique vector displacement, but it can be acted by a multitude of forces, so each force will do work along that displacement Let s start with the simplest situation: Def: If a constant force acts on an object moving in a straight line a displacement Δx, with the force making an angle θ with the direction of motion, the work done by the force on the object is given by m W x cos θ cosθ W 1 joule (J) N m Notice that, by this definition, the work is only the work done by the component of the force along the direction of motion x SI x

4 Work Comments The work describes an exchange of energy in which one object in the force-action pair looses energy and the other gains energy The nature of the work is given by its sign. Thus, work can be negative ( decreases the energy of the object), zero ( has no effect on the energy of the object), or positive ( increases the energy of the object) The sign of the work is given by cosθ in the definition: θ < 90 W >0 θ = 90 W = 0 θ > 90 W <0 v force helps the motion no effect on motion force against the motion v v Ex: As long as this person does not lift or lower the bag of groceries, he is not doing work on it, since the force P he exerts has no component in the direction of motion. Quiz: How much energy does Earth spend in a month in order to keep the Moon on the orbit?

5 Work Net work done by several forces In most cases, more than one force act on the moving objects In these cases the overall exchange of energy is measured using the net work: the sum of all individual works which is the same with the work done by the net force acting on the system. So, the calculation of the net work can be tackled in two analogous ways: 1. first find the net force and then find the work it does, or 2. first find the work done by each force and add the works Problem: 1. Net work on a flat surface: A box of mass m = 50 kg is pulled a distance d =10 m by a force =100 N making an angle = 60 o with the horizontal along a flat rough surface with coefficient of kinetic friction μ k = 0.1. Determine the net work done on the box. m θ d

6 Kinetic Energy Concept and definition How can one use the idea of work to analyze motion? According to Newton s 2 nd Law, a non-zero net force applied to an object changes its state of motion. Since the net force corresponds to a net work, we see that the work measures the change in motion In order to see how we can associate work, energy and motion, let s consider an example: a bus of mass m acted upon by a constant net force such that its acceleration a is constant and the velocity changes uniformly from v 0 to v: How is the net work done on the bus related to the change in its motion? m v 0 v W d mad v v W 2mv 2mv0 KE KE0 KE 0 2ad ad 2 v v0 So, we see that the work done by the net force results in a change of a quantity depending on mass and velocity We call this quantity kinetic energy KE d Definition: KE 1 2 mv 2

7 Kinetic Energy Relation to work Even though we derived the expression for the how the work determines a change in kinetic energy in a case with constant forces, it is a generic result: Work-Energy Theorem: The net work done by a net force on a system will lead in an equal change in the kinetic energy of the system: W KE KE KE net final initial Notice that, if the net work is positive, the kinetic energy increases, whereas, if the net work is negative, the kinetic energy decreases While the theorem is always valid (being a consequence of the conservation of energy), the definition of kinetic energy works in the form on the previous slide only for a an object in translational motion Ex: Net work on a flat surface: Recall the box in Problem 1 where we calculated the net work done on a box to be W net = 100 J. Subsequently, if we know the initial speed of the box v 0, the Work-Energy Theorem can be applied to calculate the speed v of the box after it travels the distance d: 1 1 v f k m v 0 θ KE 0 W net is done by and f k KE d W KE KE mv mv net mv mv W v v W m net net

8 Problem: 2. Work-Energy Theorem down an incline: A box of mass m = 8.00 kg is initially at rest on top of an incline of length d = 12 m and angle θ = 32. The box is released down the incline which is rough with coefficient of kinetic friction μ k = a) Calculate the net work done on the box b) Use the Work-Energy Theorem to find the velocity of the box at the base of the incline c) Use the velocity to find the acceleration, and thus confirm the formula we obtained using Newton s 2 nd law m θ

9 Potential Energy Concept and types The work done by some forces has special properties leading to a technique for solving motions using energy based on the so called potential energy If a particle is under the influence of a certain force which can virtually do work such as weight, or elastic force, but it is at rest due to a zero balance of forces, the energy is stored in the system, it is only a potential energy So, the potential energy is associated with the configuration of the system, rather than with motion. If the object is released, the potential energy can transform into kinetic energy. We ll study two types of potential energies: 1. Gravitational potential energy, PE g associated with the work done by gravity 2. Elastic potential energy, PE e associated with the work done by elastic force in springs Def: The sum between kinetic energy and potential energy in a certain configuration is called mechanical energy: E KE PE PE g e

10 Potential Energy Gravitational Consider a particle ascending vertically between two heights y 0 and y relative to an arbitrary ground, under the influence of weight and some other forces doing work W other Then, we see that the work done by gravity is: W mg y y mgy mgy PE PE PE gravity g g g y Def: gravitational potential energy: PEg mgy y 0 Therefore, the work done by gravity does not depend of the path, but only on the final and initial positions Consequently, we can rewrite the Work-Energy Theorem: KE Wnet Wgravity Wother PEg Wother KE PEg Wother KE PEg Wother mg ground This result allows us to solve problems using the concept of mechanical energy: E W E E W other 0 other mv mgy 0 gy mv m W other

11 Mechanical Energy Conservation In order to solve motion using mechanical energy, we have to write it out for different configurations of the system and then use the conservation law or instance, if the only force acting on the moving object is its weight, the mechanical energy is conserved, that is, the kinetic and potential parts must change in order to keep the total constant: Comments: mv mgy const. kinetic potential The gravitational potential energy is calculated with respect to an arbitrary ground characterized by PE g = 0. We can choose the ground at any level, since the conservation of energy involves only a difference of potential energies, which is the same in any system of coordinates. In problems, it is a good idea to choose the ground at the lowest altitude in the respective problem ground y 2 y 1 y ground y 2 y 1 y

12 Exercise: Physics wouldn t let aliens die alone While marred by multiple idea rip-offs, bad acting and a horrid script, the movie Independence Day excel at mocking elementary Physics. In the movie, translucent aliens (operating computers barely updated since late 40s) invade Earth and destroy our major cities from huge floating spaceships. At the end, these ships are destroyed by Jeff Goldblum and fall to the ground to the jubilation of happy crowds of gazers. Let s estimate how much energy would be released by dropping freely such a ship to the ground Approximate the average density of such a ship to be that of a commercial airplane (~80 kg/m 3 ). Then, simplifying one ship to be a thin cylinder of radius 12 km and height 3 km, its mass would be about kg Based on the images from the movie, the ships hover about 5 km above the ground. a) How much gravitational energy would be released by dropping such an object? b) Comment on how would this release of energy be manifested.

13 Problems: 3. Attwood machine revisited in an energy context: Two boxes with masses m 1 and m 2 are connected by a massless cord passing over a light pulley, as in the figure. When the system is released from rest (configuration A), the mass m 1 moves upward and m 2 downward a distance h with respect to the initial position. Use the conservation of mechanical energy to derive an expression for the speed of the system in configuration B. A B m 1 m 1 m 2 h h m 2

14 Elastic orce, e A variable force that we have the tools to consider is the force that appears in a deformed spring The direction and the magnitude of this force is given by Hooke s Law: If a spring is either compressed or stretched through a certain elongation x with respect to the equilibrium (un-stretched) position, the deformation will be opposed by an elastic force e proportional to the elongation: e kx where k is called the force constant and characterizes the stiffness of the spring applied e x =0 x <0 e kx equilibrium 0 L kx x >0 L 0 0 k k k applied An alternative way to write the elastic force is by using the length of the spring: if the length at equilibrium is L 0 and after elongation is L, the force is given by: e k L L 0

15 Work Done by a varying force in a 1D motion or a force that varies with position as the object moves in a straight line, calculating the work is less straightforward (needs calculus) However, in some 1D cases we can use a graphical approach: if the motion is represented on a force versus position x graph, the work is the area under the curve Ex 1: If a constant force acts parallel with the motion of an object moving through a straight displacement Δx: orce Area x 1 x 2 W x x x 2 1 x Ex 2: If a variable force (x) acts on a box moving in a straight line, such that the force has average values 1,2,3,4 along successive segments Δ x 1,2,3,4, the average work along each segment i is i Δx i which is the area of a rectangle. The sum of rectangles approximates the area under the curve when the segments Δ x are very small orce i i i (x) x 1 x2 x3 x4 1 2 W x x x x orce 3 4 (x) W = Area x

16 Potential Energy Elastic Like in the gravitational case, the work done by elastic forces depends only a final and initial configuration of the system characterized by potential energies In order to find the potential energy of a spring for a certain elongation, consider a spring compressed from an initial compression x 0 to a final one x The work will be given by the area of the graph force vs. elongation x which is simple to calculate since ~ x and the graph is a straight line: W x x kx kx PE PE PE elastic 2 e 2 e e e0 e e0 x 0 x e k k k Def: elastic potential energy: PEe 1 2 kx 2 Consequently, the Work-Energy Theorem becomes: KE W W W PE W net elastic other e other KE PE W KE PE W e other e other So, we can use the elastic potential energy as we used the gravitational one, as part of the mechanical energy e e0 0 x 0 x Work = Area x

17 Conservation of Mechanical Energy Consider an object of mass m connected to an ideal spring moving between configurations (1) and (2) characterized by vertical positions y 1,2 and spring elongations y 1,2 If there are some other forces besides gravity and elastic forces, the mechanical energy is not conserved: KE PE PE W mv kx mgy mv kx mgy W g e other other If there are no other forces, W other = 0 and the mechanical energy is conserved: mv kx mgy mv kx mgy mv 2kx mgy const E final Einitial KE PE e PE g In this expression, x is considered with respect to the equilibrium position of the spring, and y with respect to an arbitrary ground This relationship allows us to analyze motions between different configurations: just write the mechanical energy for each configuration that the system passes through and then use the energy conservation to find the relations between speeds, positions, etc.

18 Problems: 4. Conservation of mechanical energy: A 12-kg box is placed on a vertical spring of negligible mass and force constant k = 1800 N/m that is compressed 15.0 cm. When the spring is released, a) Write the mechanical energy of the system in every configuration b) What is the speed of the box when its passes through the spring s equilibrium position? c) How high does the box rise from its initial position? y 1 y 0 m k 5. Mechanical energy not conserved due to friction: A block of mass m = 0.50 kg is pushed against a horizontal spring with negligible mass and force constant k = 100 N/m, compressing it by x 0 = 0.20 m. When released, the block moves on a horizontal table top a distance x 1 =1.0 m before coming to rest. a) Write the mechanical energy of the system in every configuration b) Use conservation of energy to find the coefficient of friction between the block and the table c) Use conservation of energy to find the speed of the block when it leaves the spring m k x 0 x 1

19 Power A transfer of energy specified using the concept of work doesn t provide any information about the rate of which the energy is transferred In order to specify how fast the energy can be released or consumed by a system we use the concept of power Ex: The difference between walking and running up the stairs is power while the necessary energy is the same (given by the change in gravitational potential energy) when running the energy is spent faster: a larger power is necessary Def: The rate of energy exchanged (or work W done) by a system is called power. If a finite time interval Δt is considered, we calculate an average power The average power can be written in terms of the force and the average velocity: if a certain work is done onto an object by a force, the average power delivered is P P W t W t net d t net This shows that, for a given power, the larger the force a source of motion (such as an engine) is capable of, the slower it is delivered (smaller average acceleration) E t v J s watt W