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 W = F  d cos   Units: N m If the cart (above) is pulled at constant speed with an applied force of 20N at an angle of 60 0 over a distance of 15m, determine the work done by I) the applied force ii) gravity iii) friction iv) the net work (note: the friction force is the same as the horizontal component of the applied force) F A = 20N F F = 10 N  = 60 0  d = 15 m +- i) W A = F A  d cos  = (20N) (15m) cos 60 0 = +150 Nm ii) W g = F g  d cos  = mg  d cos (-90 0 ) = 0 Nm iii) W F = F F  d cos  = (10N) (15m) cos = Nm iv) W NET = F NET  d cos  = 0 Nm= W A + W F = +150 Nm + (-150 Nm) No work done when force is perpendicular to  d

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

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

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 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 1500 kg truck that decreases its speed from 40 m/s to 25 m/s? What’s the breaking force if the speed change occurs in a distance of 200 m W NET =  KE = 1/2 m (v f 2 - v i 2 ) m = 1500 kg v i = 40 m/s v f = 25 m/s  d = 200m W NET = ? F NET = ? = 1/2 (1500kg) ( (25m/s) 2 - (40m/s) 2 )= x 10 5 Nm F NET = W NET /  d = (-7.3 x 10 5 Nm) / 200m= NNote: negative means opposite the car’s motion = 1/2 m v f 2 - 1/2 m v i 2

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 (if g is -ve then +ve work is done lifting) and…  d = H Because doing work always changes energy from one form to another then….  GPE = - mg H = - mg (d f - d i ) A 409 kg pile driver falls from 20m to 5m. How much GPE does it lose??  GPE = - mg H = - mg (d f - d i ) m = 409 kg d i = 20 m d f = 5 m g = N/kg  GPE = ? = - (409kg)(-9.8 N/kg) (5m- 20m)  GPE = x 10 4 Nm = x 10 4 J

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

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 = 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 = J  GPE = 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

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 = (20N cos60 0 ) = 10N AREA (rectangle) = h x b = 10N x 15m = 150 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

Spring Potential Energy A physical expression for spring potential energy (SPE) can be derived using the work- energy theorem Consider a spring that is compressed a certain distance, x Hooke’s Law says that the applied force on a spring force increases proportionally with displacement or….. F applied = k x Conversly, the restoring force provided by the spring will be …. F spring = - k x From earlier, you saw that the area under the F-d graph represents the work done, so…. Where k is the spring constant for the spring Area = Work = 1/2 h b =1/2 F applied xBut… F applied = k x Work = 1/2 (k x) x =1/2 k x 2 Because doing work always changes energy from one form to another then….  SPE = 1/2 k x 2 = 1/2 k (d f 2 - d i 2 )

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 +  PE - W NC = 0 A 10g dart is put in a child’s dart gun and the spring (k = 200 N/m) is compressed 15cm. Determine the exit speed of the dart if there isn’t any friction in the barrel as the dart leaves. k = 100 N/m m dart = kg x i = 0 m x f = 0.15 m  x = 0.15 m v i = 0 m/s v f = ?  KE +  PE - W NC = 0 1/2m dart (v f 2 - v i 2 ) + 1/2 k (  x) 2 = 0 m dart (v f 2 ) + k (  x) 2 = 0 (v f 2 ) = k (  x) 2 / m dart v f =  (k (  x) 2 / m dart v f =  ((100N/m)(0.15m) 2 / kg = 15 m/s Note…W NC would be F fric (x f - x i ) Energy Change(magnitude)  KE = J  SPE = J - W NC If there was friction, the  KE would be reduced by this amount

Conservation of Energy A 200g toy car travels down a 40 0 incline starting from rest. The length of the ramp is 2m, and a constant friction force of 0.5N acts against the car’s motion. Determine the car’s speed at the bottom using a) dynamics and b) conservation of energy m c = 0.2 kg  d = 2m F F = N g = +9.8 m/s 2  = 40 0 v i = 0 m/s v f = ? b)  KE +  PE - W NC = 0 1/2m C (v f 2 - v i 2 ) + (- mgH) - F F  d = 0 Energy Change(magnitude)  KE = J  GPE = J - W NC = +1.0 J +   FgFg FNFN a) F NET = F g sin  + F F F = m c a = m c g sin  + F F a = (m c g sin  + F F ) / m c = (0.2kg)(9.8m/s 2 ) sin N) / 0.2kg = 3.8 m/s 2 v f =  (v i 2 + 2a  d )=  (2a  d )=  (2(3.8m/s 2 )(2m) ) = 3.9 m/s 1/2m C (v f 2 ) + (- mgH) - F F  d = 0 H =  d sin  = 2m sin 40 0 = 1.28m v f =  ( (mgH + F F  d) / 1/2m c ) v f =  ( (0.2)(9.8m/s 2 )(1.28m) + (-0.5N)(2m) 1/2 (0.2kg) v f = 3.9 m/s (check!!!)

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 = N 1 m/s = mi/h P = F av v av = (150 lb) (4.448 N/lb) (2.5 mi/h) (1 m/s / mi/h)= 746 W

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)= 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) = W ii) P (hp) = P (W) (1hp / 746 W)= 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