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Unformatted text preview: 1 Chapter 6: Conservation of Energy • Work by a Constant Force • Kinetic Energy • Potential Energy • Work by a Variable Force • Springs and Hooke’s Law • Conservation of Energy • Power 2 § 6.1 The Law of Conservation of Energy The total energy of the Universe is unchanged by any physical process. The three kinds of energy are: kinetic energy, potential energy, and rest energy. Energy may be converted from one form to another or transferred between bodies. 3 4 Energy forms in this chapter • Kinetic energy (translational motion) • gravitational energy • Elastic potential energy 5 Work • Work is an energy transfer that occurs when a force acts on an object that moves. • If there is no displacement, no work is done and no energy is transferred. Mg = 220 N h = 4.0 m W = 220 N X 4.0 m = 880 N.m 6 § 6.2 Work by a Constant Force Work is an energy transfer by the application of a force. For work to be done there must be a nonzero displacement. The unit of work and energy is the joule (J). 1 J = 1 Nm = 1 kg m 2 /s 2 . 7 It is only the component of the force in the direction of the displacement that does work. An FBD for the box at left: x y F w N θ The work done by the force F is: ( 29 x F r F W x x F ∆ = ∆ = θ cos ∆ r x θ F ∆ r x 8 The work done by the force N is: = N W The normal force is perpendicular to the displacement. The work done by gravity ( w ) is: = g W The force of gravity is perpendicular to the displacement. 9 The net work done on the box is: ( 29 ( 29 x F x F W W W W g N F net ∆ = + + ∆ = + + = θ θ cos cos 10 In general, the work done by a force F is defined as θ cos r F W ∆ = where F is the magnitude of the force, ∆ r is the magnitude of the object’s displacement, and θ is the angle between F and ∆ r (drawn tailtotail). Work is a Scalar can be positive , negative , or zero depending on the angle. 11 Example: A ball is tossed straight up. What is the work done by the force of gravity on the ball as it rises? FBD for rising ball: x y w ∆ r y mg y w W g ∆ = ° ∆ = 180 cos 12 Example: A box of mass m is towed up a frictionless incline at constant speed. The applied force F is parallel to the incline. What is the net work done on the box? θ F w N F x y θ cos sin = = = = ∑ ∑ θ θ w N F w F F y x Apply Newton’s 2 nd Law: 13 The magnitude of F is: θ sin mg F = If the box travels along the ramp a distance of ∆ x the work by the force F is θ sin cos x mg x F W F ∆ = ° ∆ = The work by gravity is ( 29 θ θ sin 90 cos x mg x w W g ∆ = ° + ∆ = Example continued: 14 Example continued: The work by the normal force is: 90 cos = ° ∆ = x N W N The net work done on the box is: sin sin net = + ∆ ∆ = + + = θ θ x mg x mg W W W W N g F 15 Example: What is the net work done on the box in the previous example if the box is not pulled at constant speed?...
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 Fall '08
 REHSE
 Conservation Of Energy, Energy, Force, Kinetic Energy, Potential Energy, Power, Work, Yi, Vmin

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