1. Work - Work (W) is defined as a force moved over a distance - Only the component of the force in the direction of motion does work  Units: N m The cart (above) is pulled at constant speed with a force of 20N over a distance of 15m. Determine the work done by the applied force if the handle is pulled a) in a straight line and b) at an angle of 53 0, a) F = 20 N  d = 15 m W = F H  d W = (20N)(15m) W = 300 Nm 3 4 553 0 37 0 b) F H / F = 3 / 5 F H = (3 / 5) F = (3 / 5) (20N) = 12 N W = F H  d = (12N)(15m) W = 180 Nm

2. Transformation of Energy When work is done, energy is transformed from one form into another Consider a planet moving in an elliptical orbit around the sun v v v v FgFg FgFg FgFg FgFg No work No energy change Work done slowing down planet Energy changes from kinetic to GPE Work done increasing the planet’s speed Energy changes from GPE to kinetic No work No energy change

3. Energy - Energy (E) is defined as the capacity to do workUnits: Joule (J) -Energy is the conceptual system for explaining how the universe works and accounting for changes in matter 1 Calorie (C) = 1 kcal = 4186 J -There are many types of energy which are divided up into mechanical and non-mechanical forms Form of Non- Mechanical Energy Associated with… Chemical Thermal Nuclear Electromagnetic bonds between atoms vibration of atoms bonds between protons and neutrons in nucleus Vibration of electric charges Form of Mechanical Energy Associated with… Kinetican object that is moving Gravitational Potential an object’s position in a gravitational field Elastic Potential stretched or compressed elastic materials Spring Potential stretched or compressed springs

4. Kinetic Energy A physical expression for kinetic energy can be derived using the work-energy theorem Consider an object that has a net force (F NET ) applied to it over a distance (  d) F NET vivi vfvf Change in motion W NET = F NET  d= m a  dBut v f 2 = v i 2 + 2 a  dSo.. a = ( v f 2 - v i 2 ) / 2  d W NET = m ( v f 2 - v i 2 )  d = 2  d or.. KE f - KE i =  KE What is the net work done on a 10 kg cart that increases its speed from 4 m/s to 15 m/s? What’s the force needed if the speed change occurs in a distance of 5 m W NET =  KE = 1/2 m (v f 2 - v i 2 ) m = 10 kg v i = 4 m/s v f = 15 m/s  d = 5 m W NET = ? F NET = ? = 1/2 (10kg) ( (15m/s) 2 - (4m/s) 2 )= 1045 Nm F NET = W NET /  d = (1045 Nm) / 5m= 209 N 1/2 m v f 2 - 1/2 m v i 2

5. Gravitational Potential Energy A physical expression for gravitational potential energy (GPE) can be derived using the work-energy theorem Consider an object that is lifted a certain height at constant speed in a constant gravitational field H F WT + - F = WT = mg W = F  d W = mg H and…  d = H Because doing work always changes energy from one form to another then….  GPE = mg H = mg (d f - d i ) A 50 kg pile driver falls from 5m to 1m. How much GPE does it lose??  GPE = mg H = mg (d f - d i ) m = 50 kg d i = 5 m d f = 1 m g = 10 N/kg  GPE = ? = (50kg)(10N/kg) (1m- 5m)  GPE = - 2000 Nm = - 2000 J Note: negative means GPE has decreased

6. Transformation of Energy A device that changes energy from one form to another is called a machine A car engine changes chemical energy into kinetic (moving car), gravitational potential energy (if car drives up a hill), and thermal energy (engine gets hot - exhaust gasses) Car Engine - Work is done by expanding gasses in a car engine cylinder pushing on the piston which is free to move Plants -Plants are natural machines. Nuclear energy in the sun is converted into radiant (EM) energy which is changed into chemical energy in the plant Work is done by molecular transport ( ionic pump) across the plant (or animal) cell

7. Conservation of Energy Conservative forces keep energy within a system (I.e. gravity) Energy cannot be created nor destroyed, only transferred from one form to another Non-conservative forces transfer energy out of a system (I.e. friction) Written as an expression…KE i + PE i + W NC = KE f + PE f Consider a car with 320 000J of KE braking on the flat with a force of 8000 N over a distance of 30m. What is the final energy of the car? KE i = 320 000J  d = +30 m F = -8000N PE i = PE f = 0 KE f = ? Energy Change(magnitude)  KE = KE f - KE i - 240 000J - W NC = + 240 000J KE i + PE i + W NC = KE f + PE f KE i + W NC = KE f KE i + F  d = KE f 320 000J + (-8000N) (30m) = KE f 80 000 J = KE f

8. Conservation of Energy What is the speed of the 50 kg jumper at B, C and D? Assume that there is no friction m = 50 kg g = 10 m/s 2 KE A = 0J PE A = 50 000 J d A = 100m d B = d D = 60m d c = 30m W NC = 0J v B = ? v C = ? v D = ? Energy Change A to B(magnitude)  KE = + 20 000 J  GPE = - 20 000 J KE i + PE i + W NC = KE f + PE f PE A = KE f + PE f m g d A = 1/2 m v f 2 + m g d f g d A = 1/2 v f 2 + g d f g d A - g d f = 1/2 v f 2  2g( d A - d f ) = v f At B: v B =  2g( d A - d B )=  2(10m/s 2 ) (100m - 60m) = 28 m/s At C: v C =  2g( d A - d C )=  2(10m/s 2 ) (100m - 30m) = 37 m/s= 28 m/s At D: same height as at B so same speed

9. Force-Displacement Graphs - How much work is done by a person pulling the cart 15m? The work done is the AREA under the applied force vs. displacement graph where the applied force is the component in the direction of motion. AREA (rectangle) = h x b = 12N x 15m = 180 Nm - How much work is done to stretch a spring in a spring scale 10cm? The work done is the AREA under the applied force vs. displacement graph AREA (triangle) =( h x b) / 2 = (25N x 0.1m) / 2 = 1.25 Nm Note: This is the same as F av  d

10. Power Power is the rate at which work is done Power (P) = Work / Time = W /  t Units: Nm / s or J/s or Watts (W) P = F av  d /  t = F av v av James Watt (1783) wanted to standardize the measure of power using something that everyone was familiar with ….. the power output of a horse. If a large draft horse can pull 150 lbs while walking at 2.5 mi/h determine how many Watts one “horsepower” represents. 1 lb = 4.448 N 1 m/s = 2.237 mi/h P = F av v av = (150 lb) (4.448 N/lb) (2.5 mi/h) (1 m/s / 2.237 mi/h)= 746 W

11. Power An engine is used to raise a 2000 lb load 200 m vertically up a mine shaft. If the load travels upwards at a constant speed of 3 m/s calculate: F av = 2000 lb v = 3 m/s  d = 200 m i) P = F av v av = (2000 lb) (4.448 N/lb) (3 m/s)= 26 688 W a)The power rating of the engine in i) Watts and ii) Horsepower Assume that the engine is 100% efficient (4.448 N = 1 lb) = 30 000 W ii) P (hp) = P (W) (1hp / 746 W)= 26 688 W (1hp / 746 W)= 36 hp= 40 hp b) What is the power rating (hp) of the engine if it is only 70% efficient? 0.7 W IN = W OUT 0.7 W IN /  t = W OUT /  t 0.7 P IN = P OUT 0.7 P IN = 36 W Therefore… P IN = 36 W / 0.7= 51 hp= 50 hp