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Transcription:

Chapter 7: Energy

Power: Sources of Energy

Tidal Power SF Bay Tidal Power Project

Main Ideas (Encyclopedia of Physics) Energy is an abstract quantity that an object is said to possess. It is not something you can directly observe. The usefulness of the concept comes from the Conservation of Energy. In predicting the behavior of objects, one uses the Conservation of Energy to keep track of the total energy and the transfer of energy between its various forms and between objects. Work is the transfer of energy from one object to another by a force from one on the other that displaces the other. Power is the rate at which energy is transferred or, the rate at which work is done. Power is the FLOW of energy.

Conservation of Energy Energy can neither be created nor destroyed. It may change in form or be transferred from one system to another. The total amount of energy in the Universe is constant and can never change. E i = E f Except for VERY brief amounts of time according to the Heisenberg Uncertainty

Ways to Transfer Energy Into or Out of A System Work transfers by applying a force and causing a displacement of the point of application of the force Mechanical Waves allow a disturbance to propagate through a medium Heat is driven by a temperature difference between two regions in space

More Ways to Transfer Energy Into or Out of A System Matter Transfer matter physically crosses the boundary of the system, carrying energy with it Electrical Transmission transfer is by electric current Electromagnetic Radiation energy is transferred by electromagnetic waves

Units Energy & Work are scalars and have units of the Joule: m 2 W = Fd N m = kg J s 2 = 1 m m 2 KE = mv 2 kg( ) 2 = kg = J s s 2 2

Work A force applied across a distance. W = FΔr cosθ Along the direction of motion ONLY! F must be parallel to the direction of motion!

Work: W = F Δr cos θ The displacement is that of the point of application of the force A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application The sign of the work depends on the direction of F relative to Δr Work is positive when projection of F onto Δr is in the same direction as the W F > 0 displacement Work is negative when the projection is in the opposite direction W f < 0

Energy States Left to their own devices, systems always seek out the lowest energy state available to them. Systems want to be at rest or in a constant state of motion. You have to do work on the Rock to roll it back up the hill. This will give the Rock Energy the potential of rolling back down Potential Energy.

Potential Energy The energy an object has due to its position in a force field. For example: gravity or electricity The Potential Energy is relative to a ground that is defined. (Potential Energy: U or PE)

Gravitational Potential Energy PE = mgh Force Distance It takes work to move the object and that gives it energy! Same change in height! PE = mgh The ground : h = 0 IMPORTANT! Either path gives the same potential energy! WHY?

Work Up an Incline The block of ice weighs 500 Newtons. How much work does it take to push it up the incline compared to lifting it straight up? Ignore friction.

Work Up an Incline Work = Force x Distance Straight up: W = Fd = 500N 3m= 1500J 3 F =? F = mgsinθ = 500N = 250N 6 W = Fd = 250N 6m= 1500J Push up: What is the PE at the top? 1500J mg = 500N An incline is a simple machine!

Simple Machines Force Multipliers Same Work, Different Force, Different Distance

Kinetic Energy The energy an object has due to its motion. KE = 1 2 mv 2 IMPORTANT! v is the TOTAL velocity and is a scalar!!!

Kinetic Energy Table 7.1, p.194

Quick Question (a) A guy pushes on a 20kg mower with a force of 80 N at an angle of 25 degrees. How much work does he doing pushing the mower 50 meters? W = FΔ rcosθ = 80Ncos 25x50m W = 3.63kJ

Work-Energy Thm: HO 1 Wnet = ΔKE A 2.0-kg particle has an initial velocity of (5i 4j) m/s. Some time later, its velocity is (7i + 3j) m/s. How much work was done by the resultant force during this time interval, assuming no energy is lost in the process? a. 17 J b. 49 J c. 19 J d. 53 J e. 27 J

A P HO 6 30 B A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P = 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B? The angle of the incline is 30 degrees. a. 27 J b. 20 J c. 24 J d. 17 J e. 37 J

YOU TRY HO2 A P B A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P between A and B? a. 2.7 J b. 1.0 J c. 2.2 J d. 1.6 J e. 3.2 J

Problem Suppose you push on a 30.0kg package initially at rest with a force of 120.0 N through a distance of 0.800m against an opposing friction of 5.00N. What is the kinetic energy of the box at the end of the 0.80 m? v 0 = 0

Problem Use Work-Energy Theorem: Wnet W = F d =ΔKE net net =ΔKE f F f d = KE KE ( ) KE F f d = ( ) = ( ) f i 120N 5N 0.8m= 92J v 0 = 0

Problem What is the final velocity of the box? KE = 92J KE = 1 2 mv 2 v = 2KE m = 292 J 30kg v = 2.48 m/ s v 0 = 0

Use Scalar Product of Two Vectors! The scalar product of two vectors is written as A. B It is also called the dot product A B = AxBx + AyBy + AzBz A. B = A B cos θ θ is the angle between A and B θ = v v A B AB 1 cos ( ) W = FΔr Becomes cosθ v v W = F Δr

Unit Vector Representation. How do you find the Work? Problem 7.7 A force F = ( 6ˆ i 2ˆ j )N acts on a particle that undergoes a displacement Δr = ( 3ˆ i + ˆ j )m Find (a) the work done by the force on the particle and (b) the angle between F and r.

Problem 7.7 a) F = ( 6ˆ i 2ˆ j )N Δr = ( 3ˆ i + ˆ j )m W = F Δ r = Fxx+ Fyy Find (a) the work done by the force on the particle and (b) the angle between F and r. = ( 6.00)( 3.00 ) N m + ( 2.00)( 1.00 ) N m = 16.0 J b) 1 F Δr θ = cos Δ F r 1 16 = cos = 36.9 (( 6.00) + ( 2.00) ) ( 3.00) + ( 1.00) ( ) 2 2 2 2

You Try HO5 If the resultant force acting on a 2.0-kg object is equal to (3i + 4j) N, what is the change in kinetic energy as the object moves from (7i 8j) m to (11i 5j) m? a. +36 J b. +28 J c. +32 J d. +24 J e. +60 J

You Try H04 A force acting on an object moving along the x axis is given by Fx = (14x 3.0x 2 ) N where x is in m. How much work is done by this force as the object moves from x = 1m to x = +2 m? a. +12 J b. +28 J c. +40 J d. +42 J e. 28 J

Work Done by a Varying Force The work done is equal to the area under the curve W = where F x x x i = Fcosθ f F dx x

F x (N) 20 You Try HO3 10 0 10 2 4 6 8 10 12 x (m) An object moving along the x axis is acted upon by a force F x that varies with position as shown. How much work is done by this force as the object moves from x =2m to x = 8 m? a. 10 J b. +10 J c. +30 J d. 30 J e. +40 J

Work Done by a Varying Force: Gravity Sun Probe The interplanetary probe is attracted to the sun by a force given by: F = 1.3x10 12 2 The negative sign indicates that the force is attractive. This is because of the way that the polar unit vectors are defined. With the origin located at the sun and the radial vector pointing towards the probe, the force of gravity acting on the probe is in the negative direction. r 22

Work Done by a Varying Force: Gravity The probe is moving away from the sun so the work done ON the probe BY the sun is slowing it down. Thus, the work should be negative. W 11 2.3x10 1.3x10 = 11 1.5 x10 2 x 22 dx F 1.3x10 = r 12 2 22 = x 1 22 1.3x10 ( ) = 3x10 10 J 2.3x10 1.5 x10 11 11 Attractive force versus distance for interplanetary probe. The area under the curve is negative since curve is below x-axis.

Work Done by a Varying Force Hooke s Law F s = - kx The restoring force exerted by the spring is F s = - kx x is the position of the block with respect to the equilibrium position (x = 0) k is called the spring constant or force constant and measures the stiffness of the spring

Robert Hooke (1635-1703) Leading figure in Scientific Revolution Contemporary and arch enemy of Newton Hooke s Law of elasticity Worked in Physics, Biology, Meteorology, Paleontology Devised compound microscope Coined the term cell

Hooke s Law Ut tensio, sic vis - as the extension, so is the force Hooke s Law describes the elastic response to an applied force. Elasticity is the property of an object or material which causes it to be restored to its original shape after distortion. An elastic system displaced from equilibrium oscillates in a simple way about its equilibrium position with Simple Harmonic Motion.

Elastic Systems F = kx Small Vibrations

Hooke s Law F s = - kx The Restoring Force When x is positive (spring is stretched), F is negative When x is 0 (at the equilibrium position), F is 0 When x is negative (spring is compressed), F is positive

Hooke s Law It takes twice as much force to stretch a spring twice as far. The linear dependence of displacement upon stretching force: Fapplied = kx

Hooke s Law Stress is directly proportional to strain. F ( stress) = kx( strain) applied

Hooke s Law FRestoring = kx The applied force displaces the system a distance x. The reaction force of the spring is called the Restoring Force and it is in the opposite direction to the displacement.

Spring Constant k: Stiffness The larger k, the stiffer the spring Shorter springs are stiffer springs k strength is inversely proportional to the number of coils

Spring Question Each spring is identical with the same spring constant, k. Each box is displaced by the same amount and released. Which box, if either, experiences the greater net force?

Conservative Forces The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Examples: Gravity, Spring force & Electricity The work done by a conservative force on a particle moving through any closed path is zero A closed path is one in which the beginning and ending points are the same

Conservative Forces and Potential Energy Define a potential energy function, U, such that the work done by a conservative force equals the decrease in the potential energy of the system The work done by such a force, F, is x f WC = Fx dx = ΔU xi For an infinitessimal displacement: F x = du dx

Conservative Forces and Potential Energy Check F x = du dx Look at the case of a deformed spring du dx d dx 1 2 s 2 Fs = = kx = kx This is Hooke s Law Gravitational Potential & Force: du g d Fg = = ( mgy) = mg dx dx

Energy Diagrams and Stable Equilibrium: Mass on a Spring F x du = dx The x = 0 position is one of stable equilibrium Configurations of stable equilibrium correspond to those for which U(x) is a minimum. x=x max and x=-x max are called the turning points

Energy Diagrams and Unstable Equilibrium F x = 0 at x = 0, so the particle is in equilibrium For any other value of x, the particle moves away from the equilibrium position This is an example of unstable equilibrium Configurations of unstable equilibrium correspond to those for which U(x) is a maximum. Ex: A pencil standing on its end. F x du = dx

P7.47 For the potential energy curve shown, (a) determine whether the force F x is positive, negative, or zero at the five points indicated. (b) Indicate points of stable, unstable, and neutral equilibrium. (c) Sketch the curve for F x versus x from x = 0 to x = 9.5 m. F x = du dx a) F x is zero at points A, C and E; F x is positive at point B and negative at point D. F x b) A and E are unstable, and C is stable. A B C E x (m) D

Work Done by a Spring Identify the block as the system The work is the area under the Calculate the work as the block moves from x i = - x max to x f = 0 x f 0 1 Ws = Fxdx = ( ) x kx dx = kx i xmax 2 The total work done as the block moves from x max to x max is zero. 2 max

HO 10 A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1.4 kn/m). The block is pulled 8.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position? a. 4.5 J b. 2.1 J c. 6.9 J d. 6.6 J e. 4.9 J

You Try HO 9 The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? k = 2.0 kn/m The mass of t he block is 2.0 kg. a. 3.7 m/s b. 4.4 m/s c. 4.9 m/s d. 5.4 m/s e. 14 m/s k v 2.0 kg

Conservative Forces and Potential Energy Check Look at the case of a deformed spring du dx d dx 1 2 s 2 Fs = = kx = kx This is Hooke s Law F x du = dx Gravitational Potential & Force: du g d Fg = = ( mgy) = mg dx dx

Spring with an Applied Force Suppose an external agent, F app, stretches the spring The applied force is equal and opposite to the spring force F app = -F s = -(-kx) = kx Work done by F app is equal to ½ kx 2 max

Energy in a Spring What speed will a 25g ball be shot out of a toy gun if the spring (spring constant = 50.0N/m) is compressed 0.15m? Ignore friction and the mass of the spring. W Use Energy! spring =ΔKE ball 1 1 kx = mv 2 2 v 2 2 = k x m 50.0 N / m v= (.15 m) = 6.7 m/ s.025kg