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Chapter 10

Chapter 10. Energy, Work and Simple Machines. 10.1- Energy and Work. Fd =1/2mv f 2 -1/2mv i 2 Fd represents something done to the system by the outside world W= Fd Work equals a constant force exerted on an object in the direction of motion times the object’s displacement. Kinetic Energy.

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Chapter 10

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  1. Chapter 10 Energy, Work and Simple Machines

  2. 10.1- Energy and Work • Fd=1/2mvf2-1/2mvi2 • Fd represents something done to the system by the outside world • W=Fd • Work equals a constant force exerted on an object in the direction of motion times the object’s displacement

  3. Kinetic Energy • Energy- the ability of an object to produce a change in itself or the world around it • Kinetic Energy- energy resulting from motion • KE=1/2mv2

  4. Work-energy theorem- states when work is done on an object, the result is a change in kinetic energy • W=∆KE • Work is equal to the change in kinetic energy

  5. James Prescott Joule- 19th century physicist in which the unit of energy is named after • Joule- unit of energy • 1kg m2/s2

  6. REVIEW POINT: Remember the system is the object of interest. The external world is everything else.

  7. Energy transfer can go two ways: • 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.

  8. Calculating Work • W=Fd can only be used for constant forces exerted in the direction of the motion.

  9. Example: Sun on Earth • 1st the perpendicular force does not change the speed of an object, only its direction • Since speed is constant so is its KE • KE=0 so W=0 • If F and d are at right angles F=d

  10. The unit of work is __________. • 1 __________ of work is doen when a force of 1 N acts on an object over a displacement of 1 m.

  11. Constant Force Exerted at an Angle • The work you do when you exert a force on an object, at an angle to the direction of motion, is equal to the component of the force in the direction of the displacement, multiplied by the distance moved

  12. W=FdcosӨ • Work is equal to the product of dorce and displacement, times the cosine of the angle between the force and the direction of the displacement.

  13. Steps to Solving Work-related Problems • Sketch the system and show the force that is doing the work • Draw the force and displacement vectors of the system • Find the angle, Ө, between each force and displacement • Calculate the net work done. Check the sign of the work using the direction of energy transfer. If the energy of the system has increased, the work done by that force is positive. If the energy has decreased, then the work done by that force is negative

  14. Example Problem #1 • A 105 g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. • How much work does the player do on the puck? • What is the change in energy of the system?

  15. Finding Work Done When Forces Change • If you do not have a constant force you can obtain work GRAPHICALLY.

  16. Work Done by Many Forces • If several forces are exerted on a system, calculate the work done by each force an then add the results.

  17. Power- the work done, divided by the time taken to do the work • P=W/t • P=Fd/t • P=F___ • Watt-unit of power • 1 watt=1J of energy transferred in 1 second • 1kw= 1000 watts

  18. 10.2- Machines • Machine- eases the load by changing either the magnitude or the direction of a force to match the force to the capability of the machine or person.

  19. 6 Types of Simple Machines

  20. Input work (Wi)- work you do • Output work (Wo)- work the machine does • Output work is never __________ than input work

  21. Mechanical Advantage • Effort force (Fe)- force exerted by a person on a machine • Resistance force (Fr)- force exerted by the machine • Mechanical Advantage (MA)- ratio of resistance force to effort force • MA=Fr/Fe

  22. Ideal Mechanical Advantage • Ideal Mechanical Advantage (IMA)- equal to the displacement of the effort force, divided by the displacement of the load • IMA=de/dr • The ideal mechanical advantage of an ideal machine is equal to the displacement of the effort force, divided by the displacement of the load.

  23. Efficiency • Efficiency (e)- the ratio of output work to input work • e=Wo/Wi x100 • Wo/Wi=Frdr/Fede • e=MA/IMA x100

  24. Compound Machines • Consists of 2 or more simple machines linked in a way where the resistance force of on is used as the effort force of another

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