Example problem If it takes 4 00 J of work to stretch a Hooke s law spring 10 0 cm from its unstretched length determine the extra work required to stretch it an additional 10 0 cm Conservative and Non conservative Forces Conservative Force a force for which the work it does on an object does not depend on the path Gravity is an example We know we can obtain the work from the relation W F cos d If the force is conservative then W Wc and this work can be related to the change in potential energy W c U y m mg d A A B C h mg B d d mg C W F cos d mgh h W mgsin d mgsin mgh sin W mgh Non conservative Force a force for which the work done depends on the path friction air resistance If both conservative and non conservative forces act on an object the work energy theorem is modified Wtotal WC WNC K f K i WNC K f K i WC For the case of gravity Wg WC mg yi y f WNC K f K i mg yi y f WNC K f K i mgyi mgy f WNC K U K f Ki U f U i K f U f Ki U i WNC E f Ei If no net non conservative forces WNC 0 E f Ei E Then conservation of mechanical energy holds K U Crate on Incline Revisited N fk h mg fk d x FBD N mg d WN 0 W g mgsin d mgh WC W f k mgcos d W NC The crate starts from rest vi 0 K i 0 E i U i mgh W g W NC E f E i E f E i W NC mgh K mgcos d E f Ei Some energy WNC is loss from the system In this case it is due to the non conservative friction force energy loss in the form of heat Because of friction the final speed is only 9 3 m s as we found earlier If the incline is frictionless the final speed would be E f Ei since K i 0 U f 0 2 1 2 mv f mgh Wg 2Wg 2 7510 J vf 12 3 ms m 100 kg Because of the loss of energy due to friction the final velocity is reduced It seems that energy is not conserved Conservation of Energy There is an overall principle of conservation of energy Unlike the principle of conservation of mechanical energy which can be broken this principle can not It says The total energy of the Universe is has always been and always will be constant Energy can neither be created nor destroyed only converted from one form to another So far we have only been concerned with mechanical energy There are other forms of energy heat electromagnetic chemical nuclear rest mass Em mc2 E E mech others E E Ef Ef others others E E Ei E f E E WNC others WNC E Ef Q total i mech i mech f mech f total f others i mech i mech i others i Q WNC is the energy lost or gained by the mechanical system The electrical utility industry does not produce energy but merely converts energy U mgh Lake Light heat electricity Example Problem River Hydro power plant h K mv2 2 A ball is dropped from rest at the top of a 6 10 m tall building falls straight downward collides inelastically with the ground and bounces back The ball loses 10 0 of its kinetic energy every time it collides with the ground How many bounces can the ball make and still reach a window sill that is 2 44 m above the ground Solution Method since the ball bounces on the ground there is an external force Therefore we can not use conservation of linear momentum An inelastic collision means the total energy is not conserved but we know by how much it is not conserved On every bounce 10 of K is lost Qi 0 1K i i 1 2 3 n n is the number of bounces Given h0 6 10 m hf 2 44 m Eo U o mgho E1 K1 Eo o 3 6 Since energy is conserved from point 0 to point 1 However between point 1 and 2 energy is lost Q2 0 1K1 WNC Q E final Einitial Q2 E2 E1 0 1K1 E2 E1 0 1Eo E2 Eo E2 0 9 Eo E3 E4 1 24 5 Total energy after one bounce By the same reasoning E5 0 9 E4 0 9 E2 0 9 0 9 Eo 2 E5 0 9 Eo Total energy after two bounces The total energy after n bounces is then n log h f ho E f 0 9 Eo n n log 0 9 mgh f 0 9 mgho log 2 44 6 10 n n h f 0 9 ho log 0 9 n 0 9 h f ho n 8 7 n log 0 9 log h f ho Answer is 8 n log 0 9 log h f ho bounces Power Average power Pavg or W t Units of J s Watt W Measures the rate at which work is done F cos d Pavg F cos v avg t Instantaneous power W P lim t 0 t Fd d P Fd v t W can also be replaced by the total energy E So that power would correspond to the rate of energy transfer Example A car accelerates uniformly from rest to 27 m s in 7 0 s along a level stretch of road Ignoring friction determine the average power required to accelerate the car if a the weight of the car is 1 2x104 N and b the weight of the car is 1 6x104 N Solution Given vi 0 vf 27 m s t 7 0 s a mg 1 2x104 N b mg 1 6x104 N Method determine the acceleration Pavg x F W F cos x t t We don t know the displacement x The car s motor provides the force F to accelerate the car F and x point in same direction Pavg Fx x max x t t Need ax and x v 2f v 2i 2ax x ax x Pavg v 2f v 2i 2 2 2 2 v v m m 27 0 f i 52m t 2 7 0 s 2 4 Pavg Pavg 52mg 52 1 2x10 4 6 4x10 W g 9 80 4 52 1 6x10 4 8 5x10 W b 9 80 Or from work energy theorem 1 2 2 f 2 f W K mv W mv Pavg t 2 t Same as on previous slide a
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