Conservation of Energy Energy is Conserved!

The total energy (in all forms) in a “closed” system remains constant The total energy (in all forms) in a “closed” system remains constant This is one of nature’s “conservation laws” This is one of nature’s “conservation laws” Conservation applies to: Conservation applies to: Energy (includes mass via E = mc 2 ) Energy (includes mass via E = mc 2 ) Momentum Momentum Angular Momentum Angular Momentum Electric Charge Electric Charge Conservation laws are fundamental in physics, and stem from symmetries in our space and time Conservation laws are fundamental in physics, and stem from symmetries in our space and time

Conversion of Energy A Falling object converts gravitational potential energy into kinetic energy A Falling object converts gravitational potential energy into kinetic energy Friction converts kinetic energy into vibrational (thermal) energy Friction converts kinetic energy into vibrational (thermal) energy makes things hot (rub your hands together) makes things hot (rub your hands together) it is an irretrievable energy it is an irretrievable energy Doing work on something changes that object’s energy by amount of work done, transferring energy from the agent doing the work Doing work on something changes that object’s energy by amount of work done, transferring energy from the agent doing the work

Energy Conservation Need to consider all possible forms of energy in a system e.g: Need to consider all possible forms of energy in a system e.g: Kinetic energy (1/2 mv 2 ) Kinetic energy (1/2 mv 2 ) Potential energy (gravitational mgh, electrostatic) Potential energy (gravitational mgh, electrostatic) Electromagnetic energy Electromagnetic energy Work done on the system Work done on the system Heat (1 st law of thermodynamics of Lord Kelvin) Heat (1 st law of thermodynamics of Lord Kelvin) Friction  Heat Friction  Heat Energy can neither be created nor destroyed Energy can be converted from one form to another Energy is measured in Joules [J]

Conservation of Energy Conservative forces: Gravity, electrical … Non-conservative forces: Friction, air resistance… Non-conservative forces still conserve energy! Energy just transfers to thermal energy

A small child slides down the four frictionless slides A–D. Each has the same height. Rank in order, from largest to smallest, her speeds v A to v D at the bottom. A. v A = v B = v C = v D B. v D > v A > v B > v C C. v D > v B > v C > v C D. v C > v A > v B > v D E. v C > v B > v A > v D

Conservation of Energy Three identical balls are thrown from the top of a building with the same initial speed. Initially, Ball 1 moves horizontally. Ball 2 moves upward. Ball 3 moves downward. Neglecting air resistance, which ball has the fastest speed when it hits the ground? A. Ball 1B. Ball 2C. Ball 3D. they have the same speed.

Falling (elastic) ball a E P grav a E P grav b E P grav + E K b E P grav + E K c E K c E K d E K + E P elastic d E K + E P elastic e E P elastic e E P elastic f E K f E K g E P grav g E P grav

Conservation of Energy Energy Conservation in a Pendulum Initially the pendulum has E p grav Then some E p grav is changed to E k but still retains some E p grav At the lowest point all the E p grav is now E k Which then converts back to all E p grav at the top.

Energy Conservation Demonstrated Roller coaster car lifted to initial height (energy in) Roller coaster car lifted to initial height (energy in) Converts gravitational potential energy to motion Converts gravitational potential energy to motion Fastest at bottom of track Fastest at bottom of track Re-converts kinetic energy back into potential as it climbs the next hill Re-converts kinetic energy back into potential as it climbs the next hill

Conservation of Energy Total Energy Total Energy = E P grav + E K

Energy Conservation The kinetic energy for a mass in motion is The kinetic energy for a mass in motion is E K = ½mv 2 Book dropped from rest at a height h ( E P grav = mgh) the book hits the ground with speed v. Expect ½mv 2 = mgh  s = h = ut + ½gt 2  s = h = ut + ½gt 2 where u = 0 v = u + gt  v 2 = g 2 t 2 v = u + gt  v 2 = g 2 t 2 mgh = mg  (½gt 2 ) = ½mg 2 t 2 = ½mv 2 sure enough! mgh = mg  (½gt 2 ) = ½mg 2 t 2 = ½mv 2 sure enough! Book has converted its available gravitational potential energy into kinetic energy: the energy of motion Book has converted its available gravitational potential energy into kinetic energy: the energy of motion

Loop-the-Loop In the loop-the-loop (like in a roller coaster), the velocity at the top of the loop must be enough to keep the train on the track: In the loop-the-loop (like in a roller coaster), the velocity at the top of the loop must be enough to keep the train on the track: v 2 /r > g Works out that train must start ½ r higher than top of loop to stay on track, ignoring frictional losses Works out that train must start ½ r higher than top of loop to stay on track, ignoring frictional losses ½r½r r

Conservation of Energy: Potential and Kinetic 1.What is the total energy of the sledder (m = 50 kg) at the top of the hill? 2.What is the total energy, gravitational potential energy and the kinetic energy on top of the 15 m bump? Speed? 3.What is the total energy, gravitational potential energy and the kinetic energy at the bottom of the hill? Speed? 4.How much work was done on the sledder when he was pulled up the hill by his brother?

Conservation of Energy A skier slides down the frictionless slope as shown. What is the skier’s speed at the bottom? H=40 m L=250 m start finish 28.0 m/s

x Conservation of Energy b) Find the maximum distance d the block travels up the frictionless incline if the incline angle θ is 25°. A 0.50-kg block rests on a horizontal, frictionless surface as in the figure; it is pressed against a light spring having a spring constant of k = 800 N/m, with an initial compression of 2.0 cm. a) After the block is released, find the speed of the block at the bottom of the incline, position (B). a) 0.8 m/sb) 7.7 cm

Conservation of Energy E initial = E final

Example 5.2 A diver of mass m drops from a board 10.0 m above the water surface, as in the Figure. Find his speed 5.00 m above the water surface. Neglect air resistance. 9.9 m/s

Two blocks, A and B (m A =50 kg and m B =100 kg), are connected by a string as shown. If the blocks begin at rest, what will their speeds be after A has slid a distance s = 0.25 m? Assume the pulley and incline are frictionless. s Example m/s

Example 5.7 A spring-loaded toy gun shoots a 20-g cork 10 m into the air after the spring is compressed by a distance of 1.5 cm. a) What is the spring constant? b) What is the maximum acceleration experienced by the cork? a) 17,440 N/m b) m/s 2

Mechanical Energy Conservation Re-arranging the equation slightly gives: Re-arranging the same equation slightly differently, we can write: We define Mechanical Energy to be the sum of kinetic and Potential energy:

We define Mechanical Energy to be the sum of kinetic and Potential energy: Mechanical Energy Conservation (cont) For the gravitational force (conservative force), we then have:

Example: the tallest roller coaster The tallest and fastest roller coaster is now the Steel Dragon in Japan. It has a vertical drop of 93.5 meters. At the top of the drop, cars have a velocity of 3 m/s. What is the speed of the car at the bottom of the drop (neglecting friction and air resistance)?

Problem: Roller coaster h y=0 v 0 =0 m g v=? Note: Assume no friction. Normal force does no work, so irrelevant. Final velocity is independent of the path taken!!

Example: Two boxes rest on frictionless ramps. One (the small box) has less mass than the other. They are released from rest and allowed to slide. Which box, if either, has the greater speed at B? Which, if either, has the greatest kinetic energy?

Energy Conversion/Conservation Example Drop 1 kg ball from 10 m Drop 1 kg ball from 10 m starts out with mgh = (1 kg)  (9.8 m/s 2 )  (10 m) = 98 J of gravitational potential energy starts out with mgh = (1 kg)  (9.8 m/s 2 )  (10 m) = 98 J of gravitational potential energy halfway down (5 m from floor), has given up half its potential energy (49 J) to kinetic energy halfway down (5 m from floor), has given up half its potential energy (49 J) to kinetic energy ½mv 2 = 49 J  v 2 = 98 m 2 /s 2  v  10 m/s ½mv 2 = 49 J  v 2 = 98 m 2 /s 2  v  10 m/s at floor (0 m), all potential energy is given up to kinetic energy at floor (0 m), all potential energy is given up to kinetic energy ½mv 2 = 98 J  v 2 = 196 m 2 /s 2  v = 14 m/s ½mv 2 = 98 J  v 2 = 196 m 2 /s 2  v = 14 m/s 10 m 8 m 6 m 4 m 2 m 0 m P.E. = 98 J K.E. = 0 J P.E. = 73.5 J K.E. = 24.5 J P.E. = 49 J K.E. = 49 J P.E. = 24.5 J K.E. = 73.5 J P.E. = 0 J K.E. = 98 J