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Objectives 1. Identify several forms of energy. 2. Calculate kinetic energy for an object. 3. Apply the work-kinetic energy theorem to solve problems. 4. Distinguish between kinetic and potential energy. 5. Classify different types of potential energy. 6. Calculate the potential energy associated with an object’s position. Kinetic energy • Kinetic energy is the energy of motion. • An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. • The equation for kinetic energy is: KE = ½ mv2 – Where KE is kinetic energy, in joules – v is the speed of the object, in m/s – m is the mass of the object, in kg • Kinetic energy is a scalar quantity. Questions 1. Which of the following has kinetic energy? a. a falling sky diver b. a parked car c. a shark chasing a fish d. a calculator sitting on a desk 2. If a bowling ball and a volleyball are traveling at the same speed, do they have the same kinetic energy? 3. Car A and car B are identical and are traveling at the same speed. Car A is going north while car B is going east. Which car has greater kinetic energy? Kinetic Energy depends on mass and speed • KE = ½ m v 2 • The equation shows that . . . • the more mass a body has • or the faster it’s moving • . . . the more kinetic energy it has. • Speed has more effect on Kinetic energy than mass. • KE is directly proportional to m, so doubling the mass doubles kinetic energy, and tripling the mass makes it three times greater. Kinetic energy Kinetic energy • KE is proportional to v 2, so doubling the speed quadruples kinetic energy, and tripling the speed makes it nine times greater. speed mass Sample Problem 5B • A 7.00 kg bowling ball moves at 3.00 m/s. How much kinetic energy does the bowling ball heave? How fast must a 2.45 g table-tennis ball move in order to have the same kinetic energy as the bowling ball? Is this speed reasonable for a table-tennis ball? Example 1 • An object moving at a constant speed of 25 meters per second possesses 450 joules of kinetic energy. What is the object's mass? Known: • KE = 450 J • v = 25 m/s Unknown: • m = ? kg Solve: KE = ½ mv2 450 J = ½ (m)(25 m/s)2 m = 1.4 kg example 2 • a. b. c. d. A cart of mass m traveling at a speed v has kinetic energy KE. If the mass of the cart is doubled and its speed is halved, the kinetic energy of the cart will be half as great twice as great one-fourth as great four times as great Example 3 • Which graph best represents the relationship between the kinetic energy, KE, and the velocity of an object accelerating in a straight line? a b c d Class work • Page 174 - #1-5 1. 2. 3. 4. 170 m/s 38.8 m/s The bullet with the greater mass; 2 to 1 2.4 J, 9.6 J; the bullet with the greater speed; 1 to 4 5. 1600 kg Work-kinetic energy theorem • The net work done on an object is equal to the change in the kinetic energy. Wnet Fnet x Wnet m ax Wnet KE v f vi 2ax 2 2 v f vi 2 ax Wnet m( Wnet 2 2 2 2 v f vi 2 ) 1 1 2 2 m vf m vi 2 2 Practice Problem #1 • A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car. Wnet = ∆KE = KEf - KEi (-8000N) • d = -312 500 0 J d = 39.1 m Practice Problem #2 • At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount. Wnet = ∆KE = KEf - KEi • The above problems have one thing in common: there is a force which does work over a distance in order to remove mechanical energy from an object. • The force acts opposite the object's motion and thus does negative work which results in a loss of the object's total amount of mechanical energy. In each situation, the work is related to the kinetic energy change. TMEi + Wext = TMEf KEi + Wext = 0 J ½ •m•vi2 + F•d•cos(180o) = 0 J F•d = ½ •m•vi2 d ~ vi2 Stopping distance is dependent upon the square of the velocity. Stopping distance and initial velocity Wnet = ∆KE Wnet = 0 - ½ •m•vi2 Ff•d = - ½ •m•vi2 • Ff = μFnorm = μmg • - μmg•d = - ½ mvi2 • d = vi2 / 2μg d ~ v i2 practice (m/s) Stopping Distance (m) 0 m/s 0 5 m/s 4m 10 m/s 4 m x 22 = 16 m 15 m/s 4 m x 32 = 36 m 20 m/s 4 m x 42 = 64 m 25 m/s 4 m x 52 = 100 m Example 5C • On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between and sled and the ice is 0.10? Class work • Page 176 practice 5C: #1-5 1. 2. 3. 4. 5. 7.8 m 21 m 5.1 m 300 N a.-190 J; b. -280 J; c. 750 J; d. 280 J; e. 7.6 m/s Potential energy • An object can store energy as the result of its position. Potential energy is the stored energy of position possessed by an object. • Two form: – Gravitational – Elastic Gravitational potential energy • Gravitational potential energy is the energy stored in . (height) an object as the result of its vertical position • The energy is stored as the result of the gravitational attraction of the Earth for the object. PEgrav m g h Gravitational attraction between Earth and the object: m: mass, in kilograms g: acceleration due to gravity = 9.81 m/s2 height, in meters GPE and gravity • When an object falls, gravity does positive work. Object loses GPE. • When an object is raised, gravity does negative work. Object gains GPE. Wgrav m g(hi h f ) m g(h f hi ) h h f hi Wgrav m gh Change in GPE only depends on change in height, not path As long as the object starts and ends at the same height, the object has the same change in GPE because gravity does the same amount of work regardless of which path is taken. example • The diagram shows points A, B, and C at or near Earth’s surface. As a mass is moved from A to B, 100. joules of work are done against gravity. What is the amount of work done against gravity as an identical mass is moved from A to C? 100 J As long as the object starts and ends at the same height, the object has the same change in GPE because gravity does the same amount of work regardless of which path is taken. Gravitational Potential Energy is relative: To determine the gravitational potential energy of an object, a zero height position must first be assigned. Typically, the ground is considered to be a position of zero height. But, it doesn’t have to be: – It could be relative to the height above the lab table. – It could be relative to the bottom of a mountain – It could be the lowest position on a roller coaster Unit of energy • The unit of energy is the same as work: Joules • 1 joule = 1 (kg)∙(m/s2)∙(m) = 1 Newton ∙ meter • 1 joule = 1 (kg)∙(m2/s2) Work and energy has the same unit example • • • • • How much potential energy is gained by an object with a mass of 2.00 kg that is lifted from the floor to the top of 0.92 m high table? Known: m = 2.00 kg h = 0.92 m g = 9.81 m/s2 • unknown: • PE = ? J Solve: ∆PE = mg∆h ∆PE = (2.00 kg)(9.81m/s2)(0.92 m) = 18 J • The graph of gravitational potential energy vs. vertical height for an object near Earth's surface gives the weight of the object. The weight of the object is the slope of the line. Weight = 25 J/1.0 m = 25 N m = weight / g = 2.5 kg • Springs are a special instance of a device that can store elastic potential energy due to either compression or stretching. • A force is required to compress or stretch a spring; the more compression/stretch there is, the more force that is required to compress it further. • For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k). Hooke’s Law F = kx Spring force = spring constant x displacement • F in the force needed to displace (by stretching or compressing) a spring x meters from the equilibrium (relaxed) position. The SI unit of F is Newton. • k is spring constant. It is a measure of stiffness of the spring. The greater value of k means a stiffer spring because more force is needed to stretch or compress it that spring. The SI units of k are N/m. • x the distance difference between the length of stretched/compressed spring and its relaxed (equilibrium) spring. example • Determine the x in F = kx F = kx force • Spring force is directly proportional to the elongation of the spring (displacement) The slope represents spring constant: k = F / x elongation caution elongation • Sometimes, we might see a graph such as this: The slope represents the inverse of spring constant: Slope = 1/k = x / F force example • Given the following data table and corresponding graph, calculate the spring constant of this spring. example • A 20.-newton weight is attached to a spring, causing it to stretch, as shown in the diagram. What is the spring constant of this spring? example • The graph below shows elongation as a function of the applied force for two springs, A and B. Compared to the spring constant for spring A, the spring constant for spring B is 1. smaller 2. larger 3. the same Example 12A • If a mass of 0.55 kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium position, what is the spring constant? Class work • Page 441, Practice 12A #1-4 1. 2. 3. 4. a. 15 N/m; b. less stiff 320 N/m 2700 N/m 81 N Lab 14 – Hooke’s Law (1) 1. 2. 3. 4. 5. Purpose: To determine the spring constant of a given spring. Material: spring, masses, meter stick. Procedure: Hook different masses on the spring, record the force Fs (mg) and corresponding elongation x. Plot the graph of Force vs. elongation Data section: should contain colomns: force applied, elongation. – Data measured directly from the experiment. The units of measurements in a data table should be specified in column heading only. Data analysis: Graph force vs. elongation on graph paper, answer following questions: – What does the slope mean in Force vs. elongation graph? – Determine the spring constant Force (N) Elongation (m) Force (N) Force vs. elongation Elongation (m) Elastic potential energy • Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. • Elastic potential energy can be stored in – Rubber bands – Bungee cores – Springs – trampolines Elastic potential energy in a spring • Elastic potential energy is the Work done on the spring. PEs Favg x 0kx 1 PE s ( ) x ( k x) x 2 2 1 PE s k x 2 2 • • k: spring constant x: amount of compression or extension relative to equilibrium position Elastic potential energy Elastic potential energy is directly proportional to x2 elongation Example 1 • As shown in the diagram, a 0.50-meter-long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight. If 15 joules of energy are stored in the stretched spring, what is the value of the spring constant? PE = ½ kx2 15 J = ½ k (0.50 m)2 k = 120 N/m Example 2 • The unstretched spring in the diagram has a length of 0.40 meter and a spring constant k. A weight is hung from the spring, causing it to stretch to a length of 0.60 meter. In terms of k, how many joules of elastic potential energy are stored in this stretched spring? PEs = ½ kx2 PEs = ½ k(0.20 m)2 PEs = (0.020 k) J Example 3 • Determine the potential energy stored in the spring with a spring constant of 25.0 N/m when a force of 2.50 N is applied to it. Solve: Given: PEs = ½ k∙x2 Fs = 2.50 N To find x, use Fs = kx, k = 25.0 N/m (2.50 N) = (25.0 N/m)(x) Unknown: x = 0.100 m PEs = ? J PEs = ½ (25.0 N/m)(0.100 m)2 PEs = 0.125 J Example 4 • A 10.-newton force is required to hold a stretched spring 0.20 meter from its rest position. What is the potential energy stored in the stretched spring? Given: F = 10. N x = 0.20 m Unknown: PEs = ? J PEs = Favg∙x PEs = (½ F)∙x PEs = ½ (10. N)(.20 m) PEs = 1.0 J Sample Problem 5D • A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling? Class work • Page 180 – practice 5D #1-3 1. 3.3 J 2. 0.031 J 3. a. 785 J; b. 105 J; c. 0.00 J