Chapter 10 Energy, Work, and Simple Machines

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Chapter 10 Energy, Work, and Simple Machines

•  
• In this chapter you will
•  
•  Recognize that work and power describe how the
  external world changes the energy of a system.
• Relate force to work and explain how machines
  ease the load.

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Chapter 10 Sections

• Section 10.1 Energy and Work
• Section 10.2 Machines

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Section 10.1 Energy and Work

• Objectives
• Describe the relationship between work and
  energy.
• Calculate work.
• Calculate the power used.

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WORK AND ENERGY

• Conserved Properties properties that are the
  same before and after an interaction. Examples
  are energy and momentum.
• Work is the product of the force and the
  objects displacement. It is equal to the
  constant force exerted on an object in the
  direction of motion times the objects
  displacement. It is the transfer of energy by
  Mechanical means. It is denoted by W. It is
  measured in Joules.
• W Fd
•  
• Energy the ability of an object to produce a
  change in itself or in its surroundings.
•  

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WORK AND ENERGY

• Kinetic Energy is equal to ½ times the mass of
  an object times the speed of the object squared.
  It is denoted by KE. It is measured in Joules.
• KE ½ mv2
• Work Energy Theorem states that work is equal
  to the change in Kinetic energy.
• W ?KE
•  
• James Prescott Joule physicist that established
  the relationship between work done and the change
  in energy that results. The unit of Energy is
  named after him.
•  
• Joule unit of Work. One Joule equals 1 Newton
  meter which equals 1 kgm2/s2 (1 J 1 Nm 1
  kgm2/s2).

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WORK AND ENERGY

• If the external world does work on a system then
  W is positive and the energy of the system
  increases.
• If the system does work on the external world
  then W is negative and the energy of the system
  decreases.

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CALCULATING WORK

• Work is done on an object only if it moves. If
  you hold an object in place you do not do any
  work. Also if you move an object at constant
  velocity at a constant height you do no work
  (because the force is up and the motion is
  sideways not in the same direction of the force).
• Work is only done when the force and
  displacement are in the same direction.
• A force displacement graph can give you a picture
  of the work done. The area under the curve of a
  force displacement graph represents the work
  done.
• If Force and Displacement are at right angles
  then W 0. 

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CALCULATING WORK

• Remember that you can replace a force by its
  component parts (the x and y parts of the Vector
  Force).
• The work you do when you exert a force at an
  angle to a motion is equal to the component of
  force in the direction of the motion times the
  distance moved.
•  
• The magnitude of the component of force F acting
  in the direction of motion is found by
  multiplying the force F by the cosine of the
  angle between F and the direction of motion.

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CALCULATING WORK

• Work at an Angle is equal to the product of the
  force and displacement times the cosine of the
  angle between the force and the direction of the
  displacement.
• W Fd cos ?
• The work done by Friction acts in the direction
  opposite that of motion. So the work done by
  Friction is Negative.
•  
• Negative work reduces the Kinetic Energy of the
  system.

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CALCULATING WORK

• Do Example Problem 1 p. 261
• W Fd W ?KE
• W 4.5(.15) .675 J ?KE
• W .675 Joules
•  
• Do Practice Problems p. 261 1-3
•  
• Do Example Problem 2 p. 262
• W Fd cos ?
• W 255(30) cos 25
• W 7650(.90631)
• W 6,933.25 J
• Do Practice Problems p. 262 4-8

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CALCULATING WORK

• Work is the area under the curve in a graph of
  Force versus Displacement.
•  
• If several forces are exerted on a system,
  calculate the work done by each force and then
  add the results.

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POWER

• Power is the work done divided by the time
  taken to do the work. It is denoted by P. It
  is measured in Watts.
• P W / t P Fd / t P Fd cos ? / t
  P Fv
•  
• Watt the unit of Power. It is equal to 1 Joule
  per second (J/s).
•  
• Power is often measured in Kilowatts since a watt
  is so small. 1 KW 1000 watts.
•  
• Do Example Problem 3 p. 264
• P W / t Fd / t
• P 12,000(9) / 15
• P 108,000 / 15
• P 7,200 Watts or 7.2 KW also 7,200
  J/s
•  
• Do Practice Problems p. 264 9-14

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POWER

• Since Displacement divided by time (d / t)
  velocity we can find Power using
• P Fv
• Do 10.1 Section Review p. 265 15-23