Energy, Work, and Simple Machines

1. Energy, Work, and Simple Machines Chapter 10

2. Objectives Chapter 10 Section 1 Chapter 10 Section 2 • Demonstrate a knowledge of the usefulness of simple machines. • Differentiate between ideal and real machines in terms of efficiency. • Analyze compound machines in terms of simple machines. • Calculate efficiencies for simple and compound machines. • Describe the relationship between work and energy. • Calculate work. • Calculate the work done by a variable force. • Calculate the power used.

3. 10 - 1 Energy and Work

4. Work Definition: • Work is the transfer of energy by mechanical means. • Work is done when a force is exerted on an object through a distance. • Work is only done if the force is applied for a distance! W = Fd Where: • W = work in joules (J) • F = force in newtons (N) • d = displacement in meters (m) • Note: A joule is equal to 1 Nm.

5. Kinetic Energy KE = ½ mv2 Where: • KE = kinetic energy in joules (J) • m = mass in kg • v = velocity in m/s Energy Definition: • The ability of an object to produce a change in itself or the world around it. Kinetic Energy • The energy resulting from motion. • Represented by KE.

6. Work-energy Theorem W = ΔKE Where: • W = work in joules (J) • ΔKE = change in kinetic energy • KEf – KEi • Measured in joules • The work-energy theorem states that when work is done on an object, the result is a change in kinetic energy. • Determined by English physicist James Prescott Joule in the 1800’s.

7. Example - Calculating Work Equations: W = ΔKE W = Fd So, ΔKE = Fd d = ΔKE / F When a club head strikes a a 46-g golf ball, the ball picks up 43 J of kinetic energy. A constant force of 2300 N is applied to the ball while the club head and ball are in contact. Over what distance is the club head in contact with the ball? Given: m = 46 g ΔKE = 43 J F = 2300 N d = ? Calculations: d = ΔKE / F = 43 J / 2300 N = 0.019 m

8. Homework p. 261 # 2+3 xkcd.com

9. Calculating Work So, how do you calculate work when a constant force is exerted at an angle? Remember: • A force exerted in the direction of motion is given by W=Fd. • A force exerted perpendicular to the direction of motion does no work.

10. Calculating Work when Force is exerted at an angle • Replace the force by it’s x and y components. • Determine which component of the force is in the direction of the object’s displacement. • Use this component of the force, in W = Fd, to calculate the work done.

11. Calculating Work – Force at an Angle 10.0° incline Example: How much work is done in pushing a tall box 15 m with a force of 4.0 x 102 N that is applied slightly upward at an angle of 10.0° from the horizontal? F Fy Fx 400 N 15 m Eqn & Calculations: W = Fd F = F in x-dir = F cos θ W = (F cos θ)d W = (400N)(cos 10)15m W = 5900 J Given: F = 400 N d = 15 m (in x-dir) W = ? *Need F in x-dir.

12. Finding Work when forces change • Draw a force vs. displacement graph. • The area under the curve is equal to the work done on the object. How can you calculate work when forces change?

13. Calculating Work How do you calculate work when there is more than one force? If several forces act on an object: • calculate the work done by each force • then add them together * use components for forces that are at an angle The work-energy theorem relates the net work done on the system to its change in energy.

14. Homework p. 262 #6-8 xkcd.com

15. Power Calculating Power Units: • Power in watts (W) • Work in Joules (J) • Time in seconds (s) • Power is the rate at which the external force changes the energy of the system. • The rate at which work is done. • Measured in watts (W) • 1 W = 1 J/s • Can also be measured in kilowatts (kW).

16. Example - Calculating Power Equations: P = W / t W = Fd so: P = Fd / t A net force of 2800 N accelerates a 1250-kg vehicle for 8.0 s. The vehicle travels 80.0 m during this time. What power output does this represent? Given: F = 2800 N m = 1250 kg t = 8.0 s d = 80.0 m P = ? Calculations: P = Fd / t P = (2800 N)(80.0 m) / 8.0 s P = 28000 W or 28 kW

17. classwork • This assignment will be checked before the end of class. • QUIZ on Chapter 10 Section 1 TOMORROW! • QUIZ topics include: • Work • Energy • Work-Energy Theorem • Power • pg. 264 • # 9-13

18. 10 - 2 Machines

19. What are the Six Simple Machines? • Inclined Plane • Wedge • Screw • Lever • Wheel and Axle • Pulley

20. Lever

21. Wheel and Axle

22. Pulley

23. Inclined Plane

24. Wedge

25. Screw

26. Benefits of Machines • transferring a force from one place to another • changing either the magnitude of a force • changing the direction of a force • changing the distance or speed of a force A machine is a device that eases the load by achieving one or more of the functions at the left.

27. Mechanical Advantage Mechanical Advantage • The mechanical advantage of a machine is the ratio of the resistance force to the effort force. Effort Force (Fe) • The force exerted by a person on a machine. • Also called input force. Resistance Force (Fr) • The force exerted by the machine. • Also called output force.

28. Mechanical Advantage MA > 1 MA = 1 MA < 1 • The resulting force from the machine is less than the applied force. • So, the machine changes the distance or direction of the force. • Ex. fishing pole • The resulting force from the machine is the same as the applied force. • So, the machine changes the distance or direction of the force. • Ex. pulley • The resulting force from the machine is more than the applied force. • So, the machine increases the force applied by the person. • Ex. crowbar

29. Ideal Mechanical Advantage • In an ideal machine, all energy put in would be transferred out, so the input work would be equal to the output work. • Therefore, for an ideal machine, the mechanical advantage is equal to the displacement of the effort force divided by the displacement of the load – this is called ideal mechanical advantage (IMA).

30. Efficiency

31. Compound Machines