#### Transcript Work and Energy

Work and Energy ©2009 by Goodman & Zavorotniy Conservation Principles The most powerful concepts in science are called "conservation principles". These principles allow us to solve problems without worrying much about the details of a process. We just have to take a snapshot of a system initially and finally; by comparing those two snapshots we can learn a lot. Conservation Principles A good example is a jar of candy. If you know that there are fifty pieces of candy at the beginning. And you know that no pieces have been taken out or added...you know that there must be 50 pieces at the end. Now, you could change the way you arrange them...move them around, whatever...but you still will have 50 pieces. 50 pieces still 50 pieces still 50 pieces Conservation Principles In that case we would say that the number of pieces of candy is conserved. That is, we should always get the same amount, regardless of how they are arranged as long as we take into account whether any have been added or taken away. 50 pieces still 50 pieces still 50 pieces Conservation Principles We also have to be clear about the system that we're talking about. If we're talking about one candy jar...we can't suddenly start talking about a different one and expect to get the same answers. We must define the system whenever we use a conservation principle. 50 pieces still 50 pieces still 50 pieces Conservation of Energy Energy is a conserved property of nature. It is not created or destroyed, so in a closed system we will always have the same amount of energy. The only way the energy of a system can change is if it is open to the outside...if energy has been added or taken away. You could ask "what is energy?" It turns out that energy is so fundamental, like space and time, that there is no good answer to that question. Conservation of Energy However, just like space and time, that doesn't stop us from doing very useful calculations with energy. We may not be able to define energy, but because it is a conserved property of nature, it's a very useful idea. Conservation of Energy If we call the amount of energy that we start with as "Eo" and the amount we end up with as "Ef" then we would say that if no energy is added to or taken away from a system that Eo = Ef It turns out that there are only two ways to change the energy of a system. One is with heat, which we won't deal with here; the other is with "Work", "W". If we define positive work as that work which increases the energy of a system our equation becomes: Eo + W = Ef Work Work can only be done to a system by an external force; a force from something that is not a part of the system. So if our system is a box sitting on a table...and I come along and push the box, I can increase the energy of the box...I am doing work to the box. Work Previously, we learned that the amount of work done to a system, and therefore the amount of energy increase that the system experiences is given by: Work = Force x Distanceparallel OR W = Fdparallel This is still valid, but we have to bring a new interpretation to that equation based on what we know about vector components. First, let's review some more basic facts. Work W = Fdparallel If the object that is experiencing the force does not move (if dparallel = 0) then no work is done: W = 0. The energy of the system is unchanged. If the object moves in the direction opposite the direction of the force (for instance if F is positive and dparallel is negative) then the work is negative: W < 0. The energy of the system is reduced. If the object moves in the same direction as the direction of the force (for instance if F is positive and dparallel is also positive) then the work is positive: W > 0. The energy of the system is increased. Units of Work and Energy W = Fdparallel This equation gives us the units of work. Since force is measured in Newtons (N) and distance is measured in meters (m) the unit of work is the Newton-meter (N-m). And since N = kg-m/s2; a N-m also equals a kg-m2/s2. However, in honor of James Joule, who made critical contributions in developing the idea of energy, the unit of energy is also know as a Joule (J). 1 Joule = 1 Newton-meter = 1 kilogram-meter2/second2 1 J = 1 N-m = 1 kg-m2/s2 Units of Work and Energy Eo + W = Ef Since work changed the energy of a system: the units of energy must be the same as the units of work. The units of both work and energy are the Joule. Force and Work v Previously, the force was either parallel, anti-parallel, or perpendicular. F Let's look at those three cases. Δx Force and Work F v W = Fdparallel = 0 Δx v F W = Fdparallel = FΔx Δx F v W = Fdparallel = -FΔx Δx Force and Work How do we interpret: W = Fdparallel in this case. v Δx Force and Work After breaking F into components that are parallel and perpendicular to the direction of motion, we can see that no Work is done by the perpendicular component, only by the parallel component. v θ Δx Force and Work W = FΔxcosθ The work done on an object by a force is the product of the magnitude of the force and the magnitude of the displacement times the cosine of the angle between them. v θ Δx Force and Work This is a similar case. We just have to find the component of force that is parallel to the object's displacement. Force and Work The interpretation is the same in this case, just determine the angle between the force and displacement and use: θ W = FΔxcosθ Δx A force F is at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the force F is: 1 A Fdcos θ - μ mgd B Fdcos θ v C -μ mg D 2Fdsin θ - μ mgd E Fdcos θ - 1 θ A force F is at an angle θ above the horizontal and is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the friction force is: 2 A Fdcos θ - μ mgd B 0 C -μd(mg - Fsinθ) D 2Fdsin θ - μ mgd E Fdcos θ - 1 v θ A force F is at an angle θ above the horizontal and is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the normal force is: 3 A Fdcos θ - μ mgd B 0 C -μ mgd D 2Fdsin θ - μ mgd E Fdcos θ - 1 v θ A force F is at an angle θ above the horizontal and is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the gravitational force is: 4 A Fdcos θ - μ mgd B 0 C -μ mgd D 2Fdsin θ - μ mgd E Fdcos θ - 1 v θ A force F is at an angle θ above the horizontal and is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the net force is: 5 A Fdcos θ - μ mgd B 0 C -μ mgd D 2Fdsin θ - μ mgd E Fdcos θ - 1 v θ 6 A 4 kg ball is attached to a 1.5 m long string and whirled in a horizontal circle at a constant speed of 5 m/s. How much work is done on the ball during one period? A 9J B 4.5 J C Zero D 2J E 8J Gravitational Potential Energy A book of mass "m" is lifted vertically upwards a distance "h" by an outside force. How much work does that outside force do on the book? Fapp mg W = Fdparallel Since a = 0, Fapp = mg W = (mg) dparallel Since F and d are in the same direction ...and dparallel = h W = (mg) h W = mgh Gravitational Potential Energy But we know that in general, Eo + W = Ef. If our book had no energy to begin with, Eo = 0, then W = Ef But we just showed that we did W = mgh to lift the book... so mgh = Ef Or Ef = mgh The energy of a mass is increased by an amount mgh when it is raised by a height "h". Gravitational Potential Energy The name for this form of energy is Gravitational Potential Energy (GPE). GPE = mgh One important thing to note is that changes in gravitational potential energy are important, but their absolute value is not. You can define any height to be the zero for height...and therefore zero for GPE. But whichever height you choose to call zero, changes in heights will result in changes of GPE. 7 A 2 kg block is held at the top of an incline plane. gravitational potential energy of the block? A 80 J B 60 J C 50 J D 40 J E 30 J What is the A 2 kg block released from rest from the top of an incline plane. There is no friction between the block and the surface. How much work is done by the gravitational force on the block? 8 A 80 J B 60 J C 50 J D 40 J E 30 J Kinetic Energy Imagine an object of mass "m" at rest at a height "h". If dropped, how fast will it be traveling just before striking the ground? Use your kinematics equations to get a formula for v2. v2 = vo2 + 2ad Since vo = 0, d = h, and a = g v2 = 2gh And we can solve this for "gh" gh = v2 / 2 We're going to use this result later. Kinetic Energy In this example, we dropped an object. While it was falling, its energy was constant...but changing forms. It had only gravitational potential energy, GPE, at beginning, because it had height but no velocity. It had only kinetic energy, KE, just before striking the ground, as it had velocity but no height. In between, it had some of both. Kinetic Energy Now let's look at this from an energy perspective. No external force acted on the system so its energy is constant. the form of GPE, which is "mgh". Its original energy was in Eo + W = Ef W = 0 and Eo = mgh mgh = Ef Solving for gh yields gh = Ef/m Now let's use our result from kinematics gh = v2 / 2 v2 / 2 = Ef/m Ef = 1/2 mv2 This is the energy an object has by virtue of its motion: its kinetic energy Kinetic Energy The energy an object has by virtue of its motion is called its kinetic energy. The symbol we will be using for kinetic energy is KE. Like all forms of energy, it is measured in Joules (J). The amount of KE an object has is given by: KE = 1/2 mv2 9 A D A stone is dropped from the edge of a cliff. Which of the following graphs best represents the stone's kinetic energy KE as a function of time t? C B E 10 A ball swings from point 1 to point 3. Assuming the ball is in SHM and point 3 is 2 m above the lowest point 2. Answer the following questions. What happens to the kinetic energy of the ball when it moves from point 1 to point 2? A Increases B Decreases C Remains the same D Zero E More information is required 11 A ball swings from point 1 to point 3. Assuming the ball is in SHM and point 3 is 2 m above the lowest point 2. Answer the following questions. What is the velocity of the ball at the lowest point 2? A 2.2 m/s B 3.5 m/s C 4.7 m/s D 5.1 m/s E 6.3 m/s 12 A 2 kg block released from rest from the top of an incline plane. There is no friction between the block and the surface. What is the speed of the block when it reaches the horizontal surface? A 3.2 m/s B 4.3 m/s C 5.8 m/s D 7.7 m/s E 6.6 m/s 13 A satellite with a mass m revolves around Earth in a circular orbit with a constant radius R. What is the kinetic energy of the satellite if Earth’s mass is M? A ½ mv B mgh C ½GMm/R2 D ½ GMm/R E 2Mm/R 14 An apple of mass m is thrown in horizontal from the edge of a cliff with a height of H. What is the total mechanical energy of the apple with respect to the ground when it is at the edge of the cliff? A 1/2mv02 B mgH C ½ mv02- mgH D mgH - ½ mv02 E mgH + ½ mv02 15 An apple of mass m is thrown in horizontal from the edge of a cliff with a height of H. What is the kinetic energy of the apple just before it hits the ground? A ½ mv02 + mgH B ½ mv02 - mgH C mgH D ½ mv02 E mgh - 1/2 mv02 16 A 500 kg roller coaster car starts from rest at point A and moves down the curved track. Ignore any energy loss due to friction. Find the speed of the car at the lowest point B. A 10 m/s B 20 m/s C 30 m/s D 40 m/s E 50 m/s 17 A 500 kg roller coaster car starts from rest at point A and moves down the curved track. Ignore any energy loss due to friction. Find the speed of the car when it reaches point C. A 10 m/s B 20 m/s C 30 m/s D 40 m/s E 50 m/s 18 Two projectiles A and B are launched from the ground with velocities of 50 m/s at 60 ̊ and 50 m/s at 30 ̊ with respect to the horizontal. Assuming there is no air resistance involved, which projectile has greater kinetic energy when it reaches the highest point? A Projectile A B Projectile B C They both have the same none zero kinetic energy D They both have zero kinetic energy at the highest point E More information is required 19 An object with a mass of 2 kg is initially at rest at a position x = 0. A non-constant force F is applied to the object over 6 meters. What is the total work done on the object? A 200 J B 150 J C 170 J D 190 J E 180 J 20 An object with a mass of 2 kg is initially at rest at a position x=0. A non-constant force F is applied to the object over 6 meters. What is the velocity of the object at 6 meters? A 150 m/s B 25 m/s C 300 m/s D 12.25 m/s E Not enough information 21 A D A metal ball is held stationary at a height h0 above the floor and then thrown upward. Assuming the collision with the floor is elastic, which graph best shows the relationship between the total energy E of the metal ball and its height h with respect to the floor? B E C 22 A toy car travels with speed vo at point x. Point Y is a height H below point x. Assuming there is no frictional losses and no work is done by a motor, what is the speed at point Y? A (2gH+1/2vo2)1/2 B vo-2gH C (2gH + vo2)1/2 D 2gH+(1/2vo2)1/2 E vo+2gH Elastic Potential Energy Energy can be stored in a spring, this energy is called Elastic Potential Energy. Hooke observed the relationship between the force necessary to compress a spring and how much the spring was compressed. Fspring = -kx k represents the spring constant and is measured in N/m. x represents how much the spring is compressed. The - sign tells us that this is a restorative force. Elastic Potential Energy The work needed to compress a spring is equal to the area under its force vs. distance curve. W = 1/2 (base) (height) W = 1/2 (x) (F) W = 1/2 (x) (kx) W = 1/2 kx2 Elastic Potential Energy The energy imparted to the spring by this work must be stored in the Elastic Potential Energy (EPE) of the spring: EPE = 1/2 kx2 Like all forms of energy, it is measured in Joules (J). A force of 20 N compresses a spring with spring constant 50 N/m. How much energy is stored in the spring? 23 A 2J B 5J C 4J D 6J E 8J 24 A block with a mass of m slides at a constant velocity V0 on a horizontal frictionless surface. The block collides with a spring and comes to rest when the spring is compressed to the maximum value. If the spring constant is K, what is the maximum compression in the spring? A V0 (m/k)1/2 B KmV0 C V0K/m D m V0/K E V0 (K/m)1/2 F (V0m/K)1/2 25 A block of mass m is placed on a frictionless inclined plane with an incline angle θ. The block is just in contact with a free end of an unstretched spring with a spring constant k. If the block is released from rest, what is the maximum compression in the spring? A kmgsinθ B kmgcosθ C (2mgsinθ)/k D (mg)/k E kmg Recalling conservation of energy, we can now solve more complicated problems if energy is conserved. A roller coaster is at the top of a track that is 80 m high. How fast will it be going at the bottom of the hill? Eo + W = Ef W=0 Eo = Ef E0 = GPE, Ef = KE GPE = KE Substitute GPE and KE equations mgh = 1/2 m v2 Solving for v yields v2 = 2gh Substitute the values and solve v2 = 2 (9.8) 80 This method works with many energy v =39.6 m/s problems 26 A student uses a spring (with a spring constant of 180 N/m) to launch a marble vertically into the air. The mass of the marble is 0.004 kg and the spring is compressed 0.03 m. How high will the marble go? 27 A student uses a spring gun (with a spring constant of 120 N/m) to launch a marble vertically into the air. The mass of the marble is 0.002 kg and the spring is compressed 0.04 m. How fast will it be going when it leaves the gun? 28 A student uses the lab apparatus shown below. A 5 kg block compresses a spring 6 cm. The spring constant is 300 N/m. What will the block's velocity be when released? 29 A mass is attached to an ideal spring on a smooth horizontal surface. It is displaced an amount Δxo and released. Which of the following is true? I. The KE is largest when the mass passes through the equilibrium point. II. The PE is largest when the mass has a displacement of ±Δxo. III. The PE and KE will never be equal A I only B II only C III only D I and II only E I and III only GPE and Escape Velocity The expression GPE = mgh only works near the surface of a planet. As an object goes to a great height a more general expression is needed based on Universal Gravitational Force: FGravity = GMm/r2 . Deriving this accurately requires Calculus, but you can remember it by thinking of it this way W = FΔxcosθ FGravity = GMm/r2; cosθ = -1; d = r W = -(GMm/r2)r UG = -GMm/r GPE and Escape Velocity To escape a planet, an object needs to have sufficient kinetic energy to overcome its negative gravitaional potential energy. Eo + W = Ef UG + KE + W = 0 -GMm/r + 1/2 mv2 + 0 = 0 1/2 mv2 = GMm/r v2 = 2GM/r v = (2GM/r)1/2 Note that escape velocity is independent of the mass of the escaping object. It only depends on the mass and radius of the object being escaped from. 30 A rocket of mass 5,000 kg will have _______ escape velocity as a 10,000 rocket? A larger B smaller C equal 31 Four objects are thrown with identical speeds in different directions from the top of a building. Which will be moving fastest when it strikes the ground? A B C D h E All will be the same 32 Four objects are thrown with identical speeds in different directions from the top of a building. Which will strike the ground soonest? A B C D h E All will be the same 33 Four objects are thrown with identical speeds in different directions from the top of a building. Which will go the highest? A B C D h E All will be the same 34 Four objects are thrown with identical speeds in different directions from the top of a building. Which will land farthest from the base of the building? A B C D h E All will be the same 35 Four objects are thrown with identical speeds in different directions from the top of a building. Which will have the have the greatest horizontal component of velocity at its maximum height? A B C D h E All will be the same 36 Four objects are thrown with identical speeds in different directions from the top of a building. Which will have the have the greatest kinetic energy at its maximum height? A B C h D All will be the same 37 A rocket is launched from the surface of a planet with mass M and radius R. What is the minimum velocity the rocket must be given to completely escape from the planet's gravitational field? A 2GM/16R2 B (2GM/R)1/2 C (GM/R)1/2 D 2GM/R Power It is often important to know not only if there is enough energy available to perform a task but also how much time will be required. Power is defined as the rate that work is done: P=W/t Since work is measured in Joules (J) and time is measured in seconds (s) the unit of power is Joules per second (J/s). However, in honor of James Watt, who made critical contributions in developing efficient steam engines, the unit of power is also know as a Watt (W). Power P=W/t Since W = Fd parallel P = (Fd parallel) / t Regrouping this becomes P = F(d parallel / t) Since v = d/t P = Fv parallel So power can be defined as the product of the force applied and the velocity of the object parallel to that force. Power A third useful expression for power can be derived from our original statement of the conservation of energy principle. P=W/t Since W = Ef - E0 P = (Ef - E0) / t So the power absorbed by a system can be thought of as the rate at which the energy in the system is changing. 38 A driver in a 2000 kg Porsche wishes to pass a slow moving school bus on a 4 lane road. What is the average power in watts required to accelerate the sports car from 30 m/s to 60 m/s in 9 seconds? A 1,800 W B 5,000 W C 10,000 W D 100,000 W E 300,000 W 39 A force F is applied in horizontal to a 10 kg block. The block moves at a constant speed 2 m/s across a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.5. The work done by the force F in 1.5 minutes is: A 9000 J B 5000 J C 3000 J D 2000 J E 1000 J 40 A crane lifts a 300 kg load at a constant speed to the top of a building 60 m high in 15 s. The average power expended by the crane to overcome gravity is: A 10,000 W B 12,000 W C 15,000 W D 30,000 W E 60,000 W 41 In a physics lab a student uses three light frictionless PASCO lab carts. Each cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. In which one of these three cases is more work done by force F? A cart I B cart II C cart III D the same work is done on each E more information is required 42 In a physics lab a student uses three light frictionless PASCO lab carts. Each cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. Which cart will have more kinetic energy at the end of distance d? A cart I B cart II C cart III D all three will have the same kinetic energy E more information is required In a physics lab a student uses three light frictionless PASCO lab carts. Each cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. Which cart will move faster at the end of distance d? 43 A cart I B cart II C cart III D all three will move with the same speed E more information is required 44 A box of mass M begins at rest with point 1 at a height of 6R, where 2R is the radius of the circular part of the track. The box slides down the frictionless track and around the loop. What is the ratio between the normal force on the box at point 2 to the box's weight? A 1 B 2 C 3 D 4 E 5 45 A ball of mass m is fastened to a string. The ball swings in a vertical circle of radius r with the other end of the string held fixed. Neglecting air resistance, the difference between the string's tension at the bottom of the circle and at the top of the circle is: A mg B 2mg C 3mg D 6mg E 9mg Work and Energy Calculus Based Work We learned before that work can be defined as: This equation comes from the scalar product of the force vector and displacement vector. If the force however is changing the previous equation for work does not apply. To account for the varying force like for a varying velocity when you are trying to find displacement requires the use of Calculus. Work F x By breaking the varying force into small segments we can precisely calculate the work done by a varying force. Gravitational Potential Energy Work is defined as the change in energy, so to find the potential energy due to gravity we integrate the gravitational force. Power For a constant force the power equation can be represented in several ways The average power can be represented as: To find the instantaneous power we use a limit expression in which the interval is brought to essentially zero Work along a curved path As you move from point 1 to point 2 along a curved path, the segments of the curve can be divided into an infinite amount of vector displacements denoted by dl. Φ Force and Potential Energy We already have learned that work is the change in energy. The force is dependent on the components of the potential so the force can be described as the negative gradient of the potential. Energy Diagrams An energy diagram shows the relationship between the objects position and the amount of potential energy. KE U -A 0 A -A 0 A This energy diagram shows the change of the potential for a spring. At a maximum displacement (A or -A) the object on the spring has maximum potential and at zero the object has maximum kinetic energy. Energy Diagrams Some energy diagrams are far more complex than that of a spring. To find the magnitude of the force along the curve we have to take the derivative because U unstable equilibrium stable equilibrium x Energy Diagrams In energy diagrams, stable equilibrium is anywhere the curve has a minimum and unstable equilibrium occurs when the curve has a maximum. To find the critical points take the derivative of the equation and set it equal to zero. The minimum is where the slope changes from negative to positive, and the maximum is where the slope changes from positive to negative. Energy Diagrams to Force Diagrams U U x x F F x x 46 Given the equation U(x) = x5 - 3x2 + 8. Find F(x). A 5x4 + 6x B -5x4 + 6x - 8 C -5x4 + 6x D -5x4 - 5x - 8 E 5x4 - 5x 47 Given F(x) = -24x3 + 3x2 + 8x - 7. Find U(x) from x = 0 to x = 1. A 0J B 2J C 4J D 6J E 8J A particle moves along the x-axis while acted on by a force parallel to the x-axis. The force corresponds to the potential energy function graphed. a) What is the direction of the force on the particle when it is at point A? U(J) Negative A b) What is the direction of the force on the particle when it is at point B? B C 0 Positive (m) c) What is the force on the particle when it is at point C? Zero