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Work

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Work

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  1. Work - Work (W) is defined as a force moved over a distance Units: N m - Only the component of the force in the direction of motion does work  W = F d cos  If the cart (above) is pulled at constant speed with an applied force of 20N at an angle of 600 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) - + FA = 20N FF = 10 N  = 600 d = 15 m i) WA = FA d cos  = (20N) (15m) cos 600 = +150 Nm ii) Wg = Fg d cos  = mg  d cos (-900) = 0 Nm No work done when force is perpendicular to d iii) WF = FF  d cos  = (10N) (15m) cos 1800 = - 150 Nm iv) WNET = FNET d cos  = WA + WF = +150 Nm + (-150 Nm) = 0 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 Work done slowing down planet Energychanges from kinetic to GPE v Fg Fg Fg v No work No energy change No work No energy change Fg v Work done increasing the planet’s speed Energy changes from GPE to kinetic

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

  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 (FNET) applied to it over a distance (d) Change in motion vf vi FNET FNET WNET = FNET d = m ad But vf2 = vi2 + 2 a d So.. a = ( vf2 - vi2 ) / 2d WNET = m( vf2 - vi2 )d 2 d = KE = 1/2 m vf2 - 1/2 m vi2 or.. KEf - KEi 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 m = 1500 kg vi = 40 m/s vf = 25 m/s d = 200m WNET = ? FNET = ? WNET = KE= 1/2 m (vf2 - vi2) = 1/2 (1500kg) ( (25m/s)2 - (40m/s)2) = - 7.3 x 105 Nm FNET = WNET / d = (-7.3 x 105 Nm) / 200m = - 3700 N Note: negative means opposite the car’s motion

  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 W = F d F = - WT = - mg and… d = H + F W = - mg H (if g is -ve then +ve work is done lifting) Because doing work always changes energy from one form to another then…. - WT H GPE = - mg H = - mg (df - di) A 409 kg pile driver falls from 20m to 5m. How much GPE does it lose?? m = 409 kg di = 20 m df = 5 m g = - 9.8 N/kgGPE = ? GPE = - mg H = - mg (df - di) = - (409kg)(-9.8 N/kg) (5m- 20m) GPE = - 6.0 x 104 Nm= - 6.0 x 104 J

  6. Transformation of Energy A device that changes energy from one form to another is called a machine Car Engine - 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) 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 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/s2 KEA = 0J PEA = 50 000 J dA = 100m dB = dD = 60m dc = 30m WNC = 0J vB = ? vC = ? vD = ? KEi + PEi + WNC = KEf + PEf Energy Change A to B (magnitude) KE = + 20 000 J PEA = KEf + PEf m g dA = 1/2 m vf2+ m g df g dA = 1/2 vf2+ g df g dA - g df = 1/2 vf2 GPE = - 20 000 J 2g( dA - df) = vf At B: vB = 2g( dA - dB) = 2(10m/s2) (100m- 60m) = 28 m/s At D: same height as at B so same speed At C: vC = 2g( dA - dC) = 2(10m/s2) (100m- 30m) = 37 m/s = 28 m/s

  8. 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 cos600) = 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 Favd

  9. 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….. Fapplied = k x Conversly, the restoring force provided by the spring will be …. Fspring = - k x Where k is the spring constant for the spring From earlier, you saw that the area under the F-d graph represents the work done, so…. Area = Work = 1/2 h b = 1/2 Fapplied x But… Fapplied = k x Work = 1/2 (k x) x = 1/2 k x2 Because doing work always changes energy from one form to another then…. SPE = 1/2 k x2 = 1/2 k (df2 - di2)

  10. Conservation of Energy Energy cannot be created nor destroyed, only transferred from one form to another Conservative forces keep energy within a system (I.e. gravity) Non-conservative forces transfer energy out of a system (I.e. friction) Written as an expression… KE +PE - WNC = 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 mdart = 0.010 kg xi = 0 m xf = 0.15 m  x = 0.15 m vi = 0 m/s vf = ? KE +PE - WNC = 0 KE = + 1.125 J Energy Change (magnitude) 1/2mdart(vf2 - vi2) + 1/2 k ( x)2 = 0 If there was friction, the KE would be reduced by this amount mdart(vf2 ) + k ( x)2 = 0 (vf2 ) = k ( x)2 / mdart - WNC vf= (k ( x)2 / mdart vf= ((100N/m)(0.15m)2 / 0.010 kg = 15 m/s SPE = - 1.125 J Note…WNC would be Ffric (xf - xi)

  11. Conservation of Energy A 200g toy car travels down a 400 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 + FN mc = 0.2 kg  d = 2m FF= - 0.5 N g = +9.8 m/s2 = 400vi = 0 m/s vf = ? FF a) FNET = Fg sin  + FF = mc g sin  + FF = mc a  a = (mcg sin  + FF ) / mc = (0.2kg)(9.8m/s2) sin 400 - 0.5N) / 0.2kg = 3.8 m/s2 Fg  vf =  (vi2 + 2a  d ) =  (2a  d ) =  (2(3.8m/s2)(2m) ) = 3.9 m/s b) KE +PE - WNC = 0 1/2mC(vf2 - vi2) + (- mgH) - FF  d = 0 Energy Change (magnitude) KE = + 1.5 J H =  d sin  = 2m sin 400 = 1.28m - WNC = +1.0 J 1/2mC(vf2) + (- mgH) - FF  d = 0 vf= ( (mgH + FF  d) / 1/2mc) vf= ( (0.2)(9.8m/s2)(1.28m)+(-0.5N)(2m) 1/2 (0.2kg) GPE = - 2.5 J vf= 3.9 m/s (check!!!)

  12. 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 = Favd / t = Favvav 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 = Favvav = (150 lb) (4.448 N/lb) (2.5 mi/h) (1 m/s / 2.237 mi/h) = 746 W

  13. 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: • The power rating of the engine in i) Watts and ii) Horsepower • Assume that the engine is 100% efficient (4.448 N = 1 lb) Fav = 2000 lb v = 3 m/s d = 200 m i) P = Favvav = (2000 lb) (4.448 N/lb) (3 m/s) = 26 688 W = 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 WIN= WOUT 0.7 WIN / t = WOUT / t 0.7 PIN= POUT 0.7 PIN= 36 W Therefore… PIN= 36 W / 0.7 = 51 hp = 50 hp