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Energy • Concept of energy important due to principle of conservation of energy – Energy can be converted from one form to another, but cannot be created or destroyed • There are many different kinds of energy (thermal, chemical, electromagnetic, etc.) we’ll focus on mechanical energy, specifically energy of motion (kinetic energy) • Using energy considerations, you can solve problems where you can’t use constantacceleration formulas because the applied forces aren’t constant • Definition of work: A quantity that depends on the product of an applied force and the displacement of a body while it experiences the force Work • Mathematically: – Work = (component of force in direction of displacement) (magnitude of displacement) OR – Work = (magnitude of force) (component of displacement in direction of force) • Can be summarized by: W Fx cos • Work is a scalar quantity with units of Nm = J (Joule) in SI units • Work can be positive, or zero depending negative, on the direction of F and x Sign of W 0° 90° 90° 180° + – 90° 0 Work • Pushing an eraser along a table: F f x N W • What is the sign of the work done during the displacement … – by force F on eraser: + – by friction on eraser: – – by gravitational force on eraser: 0 – by eraser on hand pushing eraser: – – Now lift the eraser. What is the sign of the work done by the gravitational force on the eraser: – • Be careful to designate work done by what force acting on what object Example Problem #5.5 Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is mk = 0.436. Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height? Partial solution (details given in class): (a) 61.3 J (b) – 46.3 J (c) 0 Total Work • When multiple forces are acting on a body, calculate the total work done on the body by: 1. Wtot = Sum of individual works due to each force acting on the body 2. Calculate net force (from vector sum) on body and use Fnet in the definition of work shown previously • Method #1 tends to be easier • Total work done on a body is related to changes in the speed of the body: – Wtot > 0: Object speeds up – Wtot < 0: Object slows down – Wtot = 0: Object maintains speed • Let’s find a quantitative relationship between Wtot and speed Total Work and Kinetic Energy • For a constant net force directed along direction of Fnet ma motion: v 2 v02 2ax • Putting these 2 equations together to eliminate a: Wtot 1 2 1 2 Fnet x mv mv0 2 2 K2 K1 • ½ mv2 has units of energy (J); we define it to be the kinetic energy K of a body (energy due to motion) – Never negative, 0 only when v = 0 • From above: Wtot K 2 K1 K (Work–Energy Theorem) – Wtot > 0 K2 > K1 v increases – Wtot < 0 K2 < K1 v decreases – Wtot = 0 K2 = K1 v unchanged CQ1: Interactive Conceptual Example: Battle of the Kinetic Energies Which block (m1 or m2) has the larger kinetic energy at the finish line? A) m1 B) m2 C) Their kinetic energies are the same at the finish line. D) Neither block has kinetic energy at the finish line. PHYSLET Exercise 7.1.3, Prentice Hall (2001) Example Problem #4.50 A car is traveling at 50.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.600? Solution is presented on the next slide. v0 Example Problem #4.50 x N v=0 f mg Wtot = K2 – K1 = ½ m(v22 – v12) = –½ mv02 (since v2 = 0) Wtot = Wf + WN + Wmg = Wf Wf = fx cos180° = –fx = – (mN)x = –mmgx –mmgx = –½ mv02 x = v02 / 2mg (a) For v0 = 50.0 km/h = 13.9 m/s, m = 0.100: x = 98.6 m (b) For m = 0.600: x = 16.4 m Potential Energy • A body gains or loses kinetic energy (energy of motion) because of work done by external forces acting on it: 1 K mv22 v12 Wtot Fx cos 2 • In some situations, it seems as if this energy were stored somewhere in a “hidden” location • Consider motion on a swing: v=0 K=0 (max height) v = vmax K = Kmax v=0 K=0 (max height) Potential Energy • It seems as though the kinetic energy is converted to some other form as you swing toward the highest points in the motion – Points to idea of energy associated with position of a body – This kind of energy is a measure of the potential (or possibility) of work to be done • Potential Energy = stored energy that can be converted to other forms of energy • Consider the fall of a book toward the floor: Gravitational Potential Energy • Once the book begins to fall, gravity does work on it, accelerating it toward the ground • Work done by gravity on the book: Wg Fg y cos Fg y mgy • More generally: Wg mg yi y f • Product of weight mg and vertical height y is defined as the gravitational potential energy: U g mgy – Has units of kgm2/s2 = J • Work done by gravity can thus be interpreted as a change in the gravitational potential energy: Wg mg y1 y2 U1 U 2 Path independent – all that matters is change in vertical position Gravitational Potential Energy • When a body moves up: – Vertical position increases relative to ground – Work done by gravity on the body is negative – Gravitational potential energy increases • When a body moves down: – Vertical position decreases relative to ground – Work done by gravity on the body is positive – Gravitational potential energy decreases • Going back to the Work – Energy Theorem: Wtot K 2 K1 • If gravity is the only force acting on the object: Wtot Wg U1 U 2 Total Mechanical Energy • Putting these 2 equations together, we get: K2 – K1 = U1 – U2 • We can rewrite this as: K1 + U1 = K2 + U2 or 1 2 1 2 mv1 mgy1 mv2 mgy2 2 2 (if only gravity does work) • Both sides of this equation must equal a constant since equation must hold for 2 arbitrary vertical points y1 (object having velocity v1) and y2 (object having velocity v2) • This constant is defined as the total mechanical energy of the system = E = K + U – “System” means the body of mass m + the earth since gravitational potential energy is a shared property of both Conservation of Mechanical Energy • Since E = constant, we say that total energy is conserved, and this is our first example of the conservation of mechanical energy • Note that conservation of total mechanical energy depends on the change in gravitational potential energy – It’s the difference in gravitational potential energy that is physically significant – Thus it doesn’t matter what height we choose to be y = 0, the origin of the vertical coordinate – We can define Ug = 0 at whatever height we choose CQ2: A ball is thrown multiple times from the top of a building with the same initial height and with the same initial speed, but with a varying launch angle relative to the horizontal. How does the speed of the ball when it hits the ground depend on the launch angle, neglecting air resistance? Projectile Launch Interactive A) It increases with increasing launch angle. B) It decreases with increasing launch angle. C) It will increase only if the ball is thrown at an angle below the horizontal. D) It does not depend on launch angle. CQ3: A 4-kg ball is thrown straight into the air at 6 m/s. How high does it travel? A) B) C) D) 1.0 m 1.6 m 1.8 m 2.0 m Design of a Loop–the–Loop Roller Coaster Suppose we wish to design the following Loop–the– 1 Loop roller coaster: 2 y 0 y1 = H R y2 = 2R What is the speed of the roller coaster car at the top of the loop (pt. 2)? (Assuming cars fall under influence of gravity only.) Conservation of mechanical energy: ½ mv12 + mgy1 = ½ mv22 + mgy2 Assume that roller coaster starts from rest at top of hill. Then we have: mgH = ½ mv22 + mg(2R) v22 = 2mg(H – 2R) / m = 2g(H – 2R) Design of a Loop–the–Loop Roller Coaster v2 = [2g(H – 2R)]1/2 (check the next time you’re at Cedar Point or King’s Island!) Work done by Varying Forces • Examples of forces that vary in magnitude with distance: – Restoring force of spring – Gravity • Example: Force directed along line of motion but varies in strength: Fx x W ≈ Fx ave,1x + Fx ave,2 x + … W = area under Fx vs. x curve Springs –x F x=0 (Spring compressed) x=0 (Spring relaxed) x x=0 F (Spring stretched) • The spring exerts a restoring force on the block: Fs kx (Hooke’s Law) Hooke's Law Interactive – k = spring constant (measures stiffness of spring) – “Slinky” has k = 1 N/m; auto suspensions have k = 105 N/m • Work done on spring to stretch: 1 2 W kx 2 • Work done by spring on block while stretching: W = –½ kx2 s F F s Springs • This expression comes from finding the area under the curve of F = kx (applied force on spring Fapp vs. x): • Avoid pitfall of saying: W = Fx = (kx)(x) = kx2 • This is wrong because F is not constant in x • Writing F = F(x) may help remind you Elastic Potential Energy • Is there potential energy stored in springs? Yes! x1 x2 F x=0 (Spring stretched to position x1) x=0 F (Spring stretched to position x2) • The work done on the block by the spring is: 1 2 1 2 W kx1 kx2 2 2 • Similar to gravity, we can define an elastic potential energy 1 2 such that W = U1,el – U2,el U el 2 kx Elastic Potential Energy Uel • Graphically: x<0 (spring compressed) 0 x x>0 (spring extended) • Note that: – Uel is > 0 always – Uel = 0 at x = 0 (relaxed position) – We do not have freedom to pick x = 0 wherever we want, in order to be consistent with U = ½ kx2 • If the elastic force is the only force that does work: Wtot = Wel = K2 – K1 = U1,el – U2,el (from the Work– Energy Theorem) Conservative and Nonconservative Forces • Rearranging terms: K1 + U1,el = K2 + U2,el • Or: ½ mv12 + ½ kx12 = ½ mv22 + ½ kx22 • In other words, the total mechanical energy E = K + U is conserved when the elastic force is the only force that does work Spring Energy Interactive • Mechanical energy is always conserved if only conservative forces are doing work (gravity, elastic forces, electromagnetic forces) – Work done by these forces are independent of path and depend only on the starting and stopping points – Can always write work done in terms of potential energy • In general, nonconservative forces are doing work as well (“Wnc”) – In these cases, the process of converting kinetic to potential energy is not reversible Conservative and Nonconservative Forces • Some nonconservative forces cause mechanical energy to be lost (“dissipative” forces) – Kinetic friction, fluid/air resistance • Other nonconservative forces cause an increase in mechanical energy – Chemical reaction in firecracker causing fragments to shoot off • In either case, we cannot write a potential energy function representative of these forces • However, total energy is always conserved if we consider the energy that changes form from one type to another • In general: K1 U1, g U1,el Wnc K 2 U 2, g U 2,el (from the Work – Energy Theorem) CQ4: A 2-kg ball and an 8-kg ball are placed on separate springs, each with the same spring constant. The springs are compressed by the same distance and released. Which of the following is true about the maximum heights reached by the balls? A) B) C) D) The 2-kg ball will go four times as high. The 2-kg ball will go twice as high. The balls will reach equal maximum heights. The 8-kg ball will go four times as high. CQ5: Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will: A) B) C) D) rise one fourth as high as object B. rise half as high as object B. rise to the same height as object B. rise twice as high as object B. Example Problem #5.39 The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in the figure. If the spring is compressed a distance of 0.120 m and the gun fired vertically as shown, the gun can launch a 20.0-g projectile from rest to a maximum height of 20.0 m above the starting point of the projectile. Neglecting all resistive forces, (a) describe the mechanical energy transformations that occur from the time the gun is fired until the projectile reaches its maximum height, (b) determine the spring constant, and (c) find the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in the figure. Partial solution (details given in class): (b) 544 N/m (c) 19.7 m/s CQ6: Interactive Example Problem: Inverse Bungee Jumper What is the necessary spring constant for the spring? A) B) C) D) 250 N/m 330 N/m 360 N/m 390 N/m ActivPhysics Online Problem 5.4 Example Problem #5.48 In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 4.0 m/s up a 20° inclined track. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move? Solution (details given in class): 1.5 m Example Problem #5.71 Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure. The 5.00-kg object is released from rest at a point 4.00 m above the floor. (a) Determine the speed of each object when the two pass each other. (b) Determine the speed of each object at the moment the 5.00-kg object hits the floor. (c) How much higher does the 3.00-kg object travel after the 5.00-kg object hits the floor? Solution (details given in class): (a) 3.13 m/s (b) 4.43 m/s (c) 1.00 m Power • Definition of work says nothing about the passage of time, but sometimes we want to know how quickly work is done • Power = time rate at which work is done (scalar quantity) • Average work done per unit time = Average power = Pave = W / t (W = work done during time interval t) • Can also write Pave = Fvave • Units of power: 1 J/s = 1 Watt (W) Elevator Power Demo CQ7: A winch is used to lift heavy objects to the top of a building under construction. A winch with a power of 50 kW was replaced with a new winch with a power of 100 kW. Which of the following statements about the new winch is NOT true? A) The new winch can do twice as much work in the same time as the old winch. B) The new winch takes twice as much time to do the same work as the old winch. C) The new winch can raise objects with twice as much mass at the same speed as the old winch. D) The new winch can raise objects with the same mass at twice the speed of the old winch.