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Chapter 5 Work and Energy

Chapter 5 Work and Energy. Definition of Work. There is a difference between the ordinary definition of work and the scientific definition of work Ordinary Definition: To do something that takes physical or mental effort

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Chapter 5 Work and Energy

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  1. Chapter 5 Work and Energy

  2. Definition of Work • There is a difference between the ordinary definition of work and the scientific definition of work • Ordinary Definition: To do something that takes physical or mental effort • Scientific definition: Work is equal to the magnitude of the applied force times the displacement of an object

  3. What is necessary for work to be done? • A force that causes displacement of an object does work on the object • Therefore, work is not done on an object unless the object is moved because of the action of a force Work is being done!!!

  4. Displacement Force What is necessary for work to be done? • Work is done only when components of a force are parallel to a displacement • When the force on an object and the object’s displacement are perpendicular, no work is done No Work was Done!!!

  5. What about forces at an angle? • Only the component of force that is parallel to the direction of the object’s displacement does work. • Example: A person pushes a box across a frictionless floor

  6. Displacement of Box Fapp,x FBD For the Box • What part of the applied force is parallel to the displacement? FN Fapp,y Fapp Fg

  7. General Equation For Work Done Work= Component of Force that does the work x displacement x cos (angle between the force vector and the displacement)

  8. If there are many constant forces acting on the object, you can find the net work done by finding the net force acting on the object Net work= Net force x displacement x cos of the angle between them Net Work done by a Constant Force

  9. Angles between vectors

  10. Units for Work • The unit for Work is the Joule • I J= 1 Nm • One Joule = One Newton x One meter

  11. The sign of work is important • Work is a scalar quantity, but it can be positive or negative • Work is negative when the force is in the direction opposite the displacement • For example, the work done by the frictional force is always negative because the frictional force is opposite the displacement

  12. Sample Problem p. 193 # 10 • A flight attendant pulls her 70 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52.0° above the horizontal. Find the following: • The work she does on the flight bag • The work done by the force of friction on the flight bag • The coefficient of kinetic friction between the flight bag and the floor

  13. FBD Fapp FN Fapp,y Ff Fapp,x Fg

  14. Fapp,x d Work done by flight attendant • Only the component of Fapp that is parallel to the displacement does work. • Fapp,x is parallel to the displacement • Fapp,x = 40cos52 = 24.63 N • Remember that in the W=Fdcosθ equation, θ represents the angle between the force vector and the displacement vector • W= (24.63 N)(253 m) cos(0)= 6230 J Θ = 0°

  15. Θ =180° d Ff Work done by friction • The bag is moving at constant velocity, so what is Ff? • Ff = Fapp,x= 40cos(52)= 24.62 N • W= Fdcosθ= (24.62N)(253m)(cos180) = -6230 J

  16. Find μk • What is FN? • Fn + Fapp,y = Fg • Fn = Fg - Fapp,y= 70N- 40sin(52)= 38.48 N

  17. I Have Kinetic Energy Energy- Section 5.2 p. 172 • Kinetic Energy- The energy of an object due to its motion I don’t Have Kinetic Energy

  18. Kinetic Energy Depends on Speed and Mass • Kinetic energy = ½ x mass x speed2 • The unit for KE is Joules (J)

  19. Sample Problem p. 173 • A 7.00 kg bowling ball moves at 3.0 m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g tennis ball move in order to have the same kinetic energy as the bowling ball?

  20. Solve the Problem • KE= ½ mv^2= ½ (7kg)(3m/s)^2= 31.5 J • How fast must 2.45 g ball move to have the same KE? • Convert g to kg  2.45 g = .00245 kg • Solve for v

  21. Work-Kinetic Energy Theorem • The net work done by a net force acting on an object is equal to the change in kinetic energy of the object • You must include all the forces acting on the object for this to work!

  22. Sample problem p. 176 #2 • A 2000 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by the traction between the wheels and the road. The other is a 950 N resistive force due to various frictional forces. Use the work-KE theorem to determine how far the car must travel for its speed to reach 2.0 m/s.

  23. What information do we have? • M= 2000 kg • Vi= 0m/s • Vf= 2 m/s Ffriction= 950 N Fforward= 1140 N

  24. What is the Work-Ke Theorem? • Remember that: • So Expand the equation to this:

  25. Solve for Fnet • Fnet= Fforward- Ffriction= 190 N forward • Rearrange the equation for d and plug in values • Why is θ= 0 in the denominator? Because the net force is in the same direction as the displacement.

  26. Potential Energy  Section 5.2 • Potential Energy is stored energy • There are two types of PE • Gravitational PE • Elastic PE

  27. Gravitational Potential Energy • Gravitational Potential Energy (PEg) is the energy associated with an object’s position relative to the Earth or some other gravitational source

  28. Elastic Potential Energy • Elastic Potential Energy (PEelastic) is the potential energy in a stretched or compressed elastic object. • k= spring (force) constant • X= displacement of spring

  29. Displacement of Spring

  30. Sample Problem p. 180 #2 • The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m?

  31. What do we know? • K = 51.0 N/m • We need to get x, in order to use the equation for elastic PE • X is the distance the spring is stretched or compressed • Relaxed length is 0.115 m, stretched length is 0.150. How much was it stretched? • 0.150- 0.115 m= 0.035 m= x

  32. Solve the problem

  33. Conservation of Energy – 5.3 • The total amount of energy in the universe is a constant • So we say that energy is conserved • From IPC: The Law of Conservation of Energy: Energy can neither be created nor destroyed

  34. Mechanical Energy • There are many types of energy (KE, PE, Thermal, etc) • We are concerned with Mechanical Energy • Mechanical Energy is the sum of kinetic energy and all forms of potential energy • ME= KE + PE

  35. Conservation of ME • In the absence of friction, mechanical energy is conserved • When friction is present, ME can be converted to other forms of energy (i.e. thermal energy) so it is not conserved.

  36. Expanded Form of Conservation of ME • Without elastic PE • With elastic PE

  37. Practice Problem p. 185 #2 • A 755 N diver drops from a board 10.0 m above the water’s surface. Find the diver’s speed 5.00 m above the water’s surface. Find the diver’s speed just before striking the water.

  38. What do we know? • W= 755 N • Initial height = 10 m • Vi= 0 m/s • There is no elastic PE involved.

  39. Solve part a. • What is the diver’s speed 5.0 m above the water’s surface? • M= Weight/g=76.96 kg • Vi= 0m/s • Initial height = 10 m • Final Height = 5 m

  40. Vi = 0 m/s Rearrange equation and solve for vf

  41. Second Part • What is the diver’s speed just before striking the water? • M= Weight/g=76.96 kg • Vi= 0m/s • Initial height = 10 m • Final Height = 0 m

  42. Finish the Problem Vi = 0 m/s hf = 0

  43. Power- Section 5.4 • Power:The rate at which work is done • The unit for power is Watts • 1 Watt = 1 J 1 s

  44. Alternate Form for Power

  45. Sample Problem (Not in book) • At what rate is a 60 kg boy using energy when he runs up a flight of stairs 10 m high in 8.0 s? • Time = 8 s • What is work done? • W=Fdcos(θ)

  46. Solve the Problem • What force does the boy apply to get himself up the stairs? • F= Weight= mg= 588.6 N • d= 10m • W= Fdcos(θ)=588.6N(10m)(cos(0)) • W=5886 J • P=W/t = 5886 J/ 8s= 735.8 Watts

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