Today's goal: I can explain and apply concepts of work and energy (3U) to real world applicaons.

Transcription

1 Lesson21.notebook September 17, 2013 Work and Energy A 3U Review Today's goal: I can explain and apply concepts of work and energy (3U) to real world applicaons. What is Work? How do we calculate it? Example: An emergency worker applies a force to push a paent horizontally for 2.44 m on a gurney with nearly friconless wheels. What is the work done if the force is applied A) horizontally with a magnitude of 15.5 N B) 25.3 o above the horizontal with a magnitude of 15.5 N.

2 Lesson21.notebook September 17, 2013 Three Requirements for Work to be Done: Situaons where work isn't done? Kinec Energy energy of moon Formula: Ex. A 55 kg sprinter has a kinec energy of 3300 J. What is the sprinter's speed?

3 Lesson21.notebook September 17, 2013 Gravitaonal Potenal Energy stored energy due to an objects elevaon above the earth's surface Formula: Ex. A diver of mass 45 kg climbs to the top of a 3.0 m diving board. How much gravitaonal potenal energy does the diver have? Conservaon of Energy Law: 1. Energy cannot be created or destroyed. 2. The total energy in a closed system always remains the same. 3. Energy can change from one form to another, but the total amount of energy stays the same as long as the system is considered closed. What does this mean?

4 Lesson21.notebook September 17, 2013 Example: A boy throws a 0.5 kg stone at 1.0 m/s towards the ground from a balcony 3.2 m above the ground. A) What is the total energy at the beginning of the throw? B) What must the energy equal when the stone hits the ground? C) What is the velocity of the stone when it hits the ground? Example: A basketball player shoots at a basket. The ball leaves the players hand at a speed of 7.2m/s from a height of 2.21 m above the ground. What is the speed of the ball as is goes through the hoop, 3.05 m above the floor? Homework: page 273, # 11 13, page 274, # 21 30

5 Lesson23.notebook September 17, 2013 Momentum and Impulse Today's goal: I can explain, in my own words, what momentum and impulse are and then solve real world problems as they arise. Momentum In your own terms: Calculang Momentum where: p = m v Which has more momentum? 1. A slow moving car or a fast moving truck? 2. A car or a truck that is travelling at the same velocity? 3. A fast moving car or a slow moving truck?

6 Lesson23.notebook September 17, 2013 Ex. What is the momentum of a 1000 kg car travelling at 80 km/h [W]? Ex. A bullet travelling at 900 m/s [W] has a momentum of 4.5 kg m/s [W]. What is its mass?

7 Lesson23.notebook September 17, 2013 Impulse the amount of force applied over a period of me Impulse is also equal to / known as the change in momentum: Ex. A baseball of mass kg is travelling at 37.5 m/s [E]. It collides with a baseball. This collision lasts for 1.15 milliseconds. Immediately aer the collision, the baseball travels at 49.5 m/s [W]. a) What is the inial momentum of the ball? b) What is the final momentum of the ball? c) What is the average force applied by the bat to the ball?

8 Lesson23.notebook September 17, 2013 Example: A 0.06 kg ball is thrown upward and struck just as it comes to rest at the top of its motion. The racket exerts an average horizontal force of magnitude 4.2 x 10 2 N on the tennis ball. A) Determine the speed of the ball after the collision is the average force is exerte for s. B) Same as part 'A', but the time lasts for s. C) Explain the meaning and advantage of follow through in this example. Homework: page 218, #23 36

9 Lesson24.notebook September 17, 2013 Conservation of Momentum In One Dimension Today's goal: I can explain the meaning of conservaon of momentum and I can apply it in real world applicaons. Law of Conservaon of Momentum The total momentum of a system before a collision is equal to the total momentum in the system aer the collision. In Easier Terms: We know that p system = p' system, but now we are dealing with multiple factors so: Example: A 75 kg man is standing on a staonary ra with mass 55 kg. He walks towards the front of the ra with a velocity of 2.3 m/s [W]. What is the resulng velocity of the ra?

10 Lesson24.notebook September 17, 2013 Example: During a football game, a fullback with a mass of 108 kg is running with a velocity of 9.1 m/s [E]. He gets tackled by a linebacker (m=91 kg) coming at him with a velocity of 6.3m/s [W]. What is the velocity of both men during the tackle? Example: A kg golf ball is hit with a driver. The head of the driver has a mass of 0.15 kg and travels at a velocity of 56 m/s before the collision and a velocity of 47 m/s directly aer the collision. What is the speed of the ball? Homework: page 219, # 37 45

11 Lesson25.notebook September 17, 2013 Conservation of Momentum In Two Dimensions Today's goal: I can explain the meaning of conservation of momentum in two dimensions and I can apply it in real world applications. The law of conservation of energy applies to any situation in which a system is subject to a net force of zero. This would include situations that contain two dimensional forces. To deal with this we fall back on our knowledge of vectors. Example: A steel ball of mass 0.5 kg, moving with a velocity of 2.0 m/s [E], strikes a tennis ball of mass 0.3kg, inially at rest. Aer they collide, the steel ball has a a velocity of 1.5 m/s [E30 o N]. What is the velocity of the tennis ball?

12 Lesson25.notebook September 17, 2013 Example: A 2 kg rock is thrown with a velocity of 9.5 m/s. The rock hits a window and breaks into two pieces. If one piece of the rock is 0.65 kg and bounces off the window with a velocity of 1.7 m/s [45 o to the horizontal], what is the velocity of the second piece of the rock? Example: A bomb sing at rest breaks into 3 pieces when it is set off. If piece 1 is 3.2 kg with a velocity of 3.4 m/s [N25 o E], piece 2 is 0.75 kg with a velocity of 9.8 m/s [E78 o S] and piece 3 has a speed of 5.3 m/s, what is the mass of the original bomb? Homework: page 220, # 47 56

13 Lesson26.notebook September 17, 2013 Hooke's Law Today's goal: I can explain and apply Hooke's Law to real world applications. The force exerted by an elastic device, is proportional to the distance the spring has moved from its equilibrium position. Diagram: Equilibrium vs. Compressed/Stretched For the "ideal spring Robert Hooke published his research that stated that F x = kx where: F x k x and if x > 0 (stretched) then F x < 0 (pull) and if x < 0 (compressed) then F x > 0 push) Analysis of Hooke's Law

14 Lesson26.notebook September 17, 2013 Example: A student stretches a spring horizontally a distance of 15mm by applying a force of 0.18 N[E]. a) Determine the force constant for the spring. b) What is the force that the spring is exerng onto you? Example: You are bungee jumping off a bridge that is 100 m above the ground. The bungee cord has a length of 60 m. What is the minimum value of k for the bungee cord that you would enable your to go bungee jumping? (Assume you are 60 kg and 1.5m tall)

15 Lesson26.notebook September 17, 2013 Elastic Potential Energy The energy stored in objects that are stretched, compressed, bent or twisted is called Elastic Potential Energy. If we again look at a the graph, Elastic Force vs. Displacement, we know the slope of the line is the force constant. What is the are under the line though? Example: An apple of mass 1.0 kg is attached to a vertical spring with force constant of 9.6 N/m. The apple is held so that the spring is at its equilibrium position, and then released and allowed to fall. Neglecting the mass of the spring: A) How much elastic potential energy is stored in the spring when the apple has fallen 11 cm? B) What is the speed of the apple when it has fallen 11 cm? Homework: page 276 # 41 50

16 Lesson27.notebook September 17, 2013 Elastic and Inelastic Collisions Today's goal: I can differentiate between elastic and inelastic collisions and I can complete the appropriate calculations for real world applications. Thus far we have discussed two types of laws of conservation: Conservation of Momentum: m 1 v 1 + m 2 v 2 = m 1 'v 1 ' + m 2 'v 2 ' Conservation of Energy: Fd + 1/2mv 2 + mgh + 1/2kx 2 = F'd' + 1/2mv' 2 + mgh + 1/2kx' 2 For each of these we are going to revisit, however we are going to analyze it from linear perspective. There are two types of collisions: Elastic and Inelastic. Elastic collisions occur when both kinetic energy and momentum are conserved. No deformation of the objects occur. Inelastic collisions occur only when momentum is conserved. Deformation or energy loss occurs. Equations for One Dimensional Elastic Collisions By combining momentum and energy the following derivation leads to shortcuts when calculating final velocities for two objects when the second object is at rest. Momentum Kinetic Energy

17 Lesson27.notebook September 17, 2013 Carts and Spring Elastic vs. Inelastic Collisions Elastic Inelastic F F x x F F x x F F F x F x x x

18 Lesson27.notebook September 17, 2013 Flow Chart Example: A 5.5 g bullet is fired at a suspended 2.0 kg block of wood. After impac the block of wood swings such that it has a height change of 13 cm above the rest position. What was the velocity of the bullet just before it hit and became embedded in the block of wood?

19 Lesson27.notebook September 17, 2013 Example: A 0.5 kg block of wood slides across the floor at 3 m/s toward a 3.5 kg block of wood. Nails are sticking out of the sides of each block causing the two blocks to stick together after the collision. If the blocks moved a total of 3.0 cm aft impact, what was the coefficient of friction between the blocks and the floor? A 400 ton train and 600 ton train are involved in an elastic collision (due to the cow bumper on each train and low velocity) on a straight section of track. The 400 ton train was travelling at 2 m/s striking the second train which is at rest. Determine th velocities of each train after the collision. Homework: page 278, # handout, All questions

20 Lesson28.notebook September 17, 2013 Simple Harmonic Motion Today's goal: I can explain, provide examples and solve problems related to simp harmonic motion. Any motion that obeys Hooke's Law is called Simple Harmonic Motion. With respect to springs, SHM occurs in the following steps: 1) An initial force stretches or compresses the spring creating E E. 2) The mass moves in the direction of the equilibrium point. 3) When the mass reaches the equilibrium point either the kinetic energy or gravitational potential carries the mass past the equilibrium point. 4) The spring reaches the maximum extension/compression length from step 1. 5) The process returns to step 2 and repeats indefinitely in an ideal system. Diagram To find a relationship we now compare Newton's Second Law and Hooke's Law: If we are dealing with a pendulum instead of spring mass system Hooke's law changes to look like: The total energy in a system can be defined by:

21 Lesson28.notebook September 17, 2013 Example: A spring with a spring constant of 80 N/m has a 1.5 kg block attached to its free end. If the block is pulled out 50 cm from the equilibrium position and released, answer the following (assume no friction): A) What is the maximum acceleration of the block? B) What is the acceleration of the block at the equilibrium position? C) What is the speed of the block as it returns to the equilibrium position? D) What is the total energy for this system?

22 Lesson28.notebook September 17, 2013 Damped Simple Harmonic Motion When damping is applied to SHM, the amplitude of each oscillation is decreased until it reaches zero. This concept is important for many applications in the world around us. There are three types of damping: 1) Overdamping: Oscillation ceases and the mass slowly returns to the equilibrium position. Examples: 2) Critical damping: Oscillations cease and the mass moves back to equilibrium position as fast as possible. This is a theoretical situation as it is never perfectly reached in nature. Examples: 3) Underdamping: Oscillation is continually reduced in amplitude. Examples: Engineering Application for Damped SHM: Skyscrappers and oscillation.

23 Lesson28.notebook September 17, 2013 Damped SHM equations become very complicated and require 2nd year university calculus to solve. However, for our purposes we can "simplify" the situation by choosing an equation that closely models the scenario and understand that it is no ideal. In grade 11 math you discussed exponential functions. There were two types: We are interested in the second type of exponential function. The E E involves this addition to the original function to create the "damping". Example: The energy for a damped SHM spring is defined as: E E = 0.5kx 2 (e bt/m) ), where the spring constant is 50 N/m, the mass is 4 kg and the decay constant, b, i 0.05 kg/s. If the spring and mass are stretched 25 cm and then released calculate A) The energy after 0.5 seconds. B) The energy after 5 minutes. C) The approximate time for the energy to reach 0 J. Homework: page 314, # 30 40