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WORK, ENERGY, AND POWER SPH4C – Unit 3 Day 2 Energy
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BELL WORK
| #1 A farmer applies a constant horizontal force of magnitude 21 N on a wagon and moves it a horizontal distance of 3.2 m. Calculate the work done by the farmer on the wagon.
| #2 A truck does 3.2 kJ of work pulling horizontally on a car to move it 1.8 m in the direction of the force. Calculate the magnitude of the force.
Wedd= 21N × 3.2 m
a 67 J
0
Ws 3.2 kJ = 3200J W = Fed
Ddt 1.8 m EWeetnoooononT.EE#k
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ZERO WORK
| If the force and displacement are perpendicular to each other, the work done by the force is zero.
| If no force is applied work is zero.
| If there is no displacement work is zero.
90°
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SOME USEFUL DEFINITIONS
| Energy y the capacity (ability) to do work
| Kinetic Energy (EK) y energy possessed by moving objects
| Work – Energy Principle y the net amount of mechanical work done on an object
equals the objects change in kinetic energy
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MORE USEFUL DEFINITIONS | Potential Energy
y a form of energy an object possesses because of its position in relation to forces in the environment
| Gravitational Potential Energy
y energy possessed by an object due to its position relative to the surface of the earth
| Reference Level y a designated level to which objects may fall y Considered to have a gravitational potential energy value of
0 J | Mechanical Energy
y the sum of kinetic energy and gravitational potential energy
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KINETIC ENERGY
| scalar quantity
| Equation:
| OR
2
2kmvE
212kE mv
Check Units
←-
R
bug D mimIN s sxs
9 =÷.
kg_m.mkg.gr#s2=1Nm
= 1J
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WORK ENERGY PRINCIPLE
| The net amount of mechanical work done on an object equals the object’s change in kinetic energy
Net kW E '
Net kf kiW E E �
2 2
2 2f i
Net
mv mvW �
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GRAVITATIONAL POTENTIAL ENERGY
gE mgh Check Units ThighFyE!k Teena .
N . m= 1J
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MECHANICAL ENERGY
mechanical K gE E E �
+Gravitational
Kinetic PotentialEnergy Energi
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EXAMPLE #1 GRAVITATIONAL POTENTIAL ENERGY
| A 0.45 kg book is resting on a desktop 0.64 m high. Calculate the books gravitational potential energy relative to the desktop and the floor.
Given MTO -45 kgo
- hoo .64m
$0 .64m ( above floor)Requited Eg ( above desk) h=0
Eg( above floor) ( above desk)Select : Egimgh Relative to floor
Eps : Relative todeskEg=¢4skg)(98s⇒¢.64m)
tEjs←8↳l.sn#dEg=2.8T
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EXAMPLE #2 GRAVITATIONAL POTENTIAL ENERGY
| The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (mass 72 kg) to the top of the hill. If the skier’s gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
Pg . 142*4
Elevations '3S0ntBo4m Steps
jI→¥¥wIHIY¥÷g↳¥on sedtsttnghhquooom
Givei_m=72kg Utterly
h.LI#Eg=9.2X10IEkjatwngh=l304m
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EXAMPLE #3 KINETIC ENERGY
| During a shot put a 5.0 kg shot leaves an athletes hand at 13 m/s. Calculate the kinetic energy.
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EXAMPLE #4 MECHANICAL ENERGY
| A roller coaster car (mass 350 kg) is moving at a speed of 2 m/s at the top of a 65 m hill. Calculate the mechanical energy of the roller coaster car.
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QUESTIONS
| Pg. 137 #11-14 | Pg. 138 #1-7 | Pg. 142-143 #1,3,5-10 | Pg. 146 #1-4,6
•