Chapter 6 Kinetic energy and work I Kinetic energy II Work III Work Kinetic energy theorem IV Work done by a constant force Gravitational force V Work done by a variable force Spring force General 1D Analysis 3D Analysis Work Kinetic Energy Theorem VI Power Energy scalar quantity associated with a state or condition of one or more objects I Kinetic energy Energy associated with the state of motion of an object K 1 2 mv 2 7 1 Units 1 Joule 1J 1 kgm2 s2 N m II Work Energy transferred to or from an object by means of a force acting on the object To W From W Constant force v 2 v02 2a x d a x Fx ma x v 2 v02 2d 1 1 1 m v 2 v02 ma x d m v 2 v02 2 d 2 1 m v 2 v02 K f K i Fx d W Fx d 2 Fx ma x Work done by the force Energy transfer due to the force 1 To calculate the work done on an object by a force during a displacement we use only the force component along the object s displacement The force component perpendicular to the displacement does zero work W Fx d F cos d F d F 7 3 cos d 90 W Assumptions 1 F cte 2 Object particle like 180 90 W Units 1 Joule 1J 1 kgm2 s2 90 0 A force does W when it has a vector component in the same direction as the displacement and W when it has a vector component in the opposite direction W 0 when it has no such vector component Net work done by several forces Sum of works done by individual forces Calculation 1 W net W 1 W 2 W 3 2 Fnet W net Fnet d II Work Kinetic Energy Theorem K K f K i W 7 4 Change in the kinetic energy of the particle Net work done on the particle III Work done by a constant force Gravitational force W F d mgd cos 7 5 Rising object W mgd cos180 mgd Fg transfers mgd energy from the object s kinetic energy Falling object W mgd cos 0 mgd Fg transfers mgd energy to the object s kinetic energy 2 External applied force Gravitational force K K f K i Wa Wg 7 6 Object stationary before and after the lift W a W g 0 The applied force transfers the same amount of energy to the object as the gravitational force transfers from the object IV Work done by a variable force Spring force F kd 7 7 Hooke s law k spring constant measures spring s stiffness Units N m Hooke s law 1D F x kx Work done by a spring force Assumptions Spring is massless mspring mblock Ideal spring obeys Hooke s law exactly Contact between the block and floor is frictionless Block is particle like Fx Calculation 1 The block displacement must be divided into many segments of infinitesimal width x xi x xf x Fj 2 F x cte within each short x segment 3 xf xf i i Ws F j x x 0 Ws x F dx x kx dx xf 1 WS k x dx k x 2 xi 2 Ws 1 1 k xi2 k x 2f 2 2 1 Ws k x 2f 2 xf xi 1 k x 2f xi2 2 W s 0 If Block ends up at xf xi if xi 0 Work done by an applied force spring force K K f K i Wa Ws Block stationary before and after the displacement K 0 W a W s The work done by the applied force displacing the block is the negative of the work done by the spring force Work done by a general variable force 1D Analysis W j F j avg x W W j F j avg x better approximation more x x 0 xf W lim F j avg x W F x dx x 0 7 10 xi Geometrically Work is the area between the curve F x and the x axis 4 3D Analysis F Fx i Fy j Fz k Fx F x Fy F y Fz F z dr dx i dy j dz k r x y z dW F dr Fx dx Fy dy Fz dz W dW Fx dx Fy dy Fz dz f f f f ri xi yi zi Work Kinetic Energy Theorem Variable force xf xf W F x dx ma dx xi xi dv ma dx m dx dt m dv v dx mvdv dx dv dv dx dv v dt dx dt dx vf vf W mv dv m v dv vi vi 1 2 1 2 mv f mvi K f K i K 2 2 V Power Time rate at which the applied force does work Average power amount of work done in an amount of time t by a force Pavg W t 7 12 Instantaneous power instantaneous time rate of doing work P P dW dt F 7 13 dW F cos dx dx F cos Fv cos F v dt dt dt x 7 14 Units 1 Watt 1 W 1J s 1 kilowatt hour 1 kW h 3 60 x 106 J 3 6 MJ 5 54 In the figure a below a 2N force is applied to a 4kg block at a downward angle as the block moves rightward through 1m across a frictionless floor Find an expression for the speed vf at the end of that distance if the block s initial velocity is a 0 and b 1m s to the right c The situation in b is similar in that the block is initially moving at 1m s to the right but now the 2N force is directed downward to the left Find an expression for the speed of the block at the end of the 1m distance W F d F cos d N Fx Fx Fy mg W K 0 5m v 2f v02 N Fy mg a v0 0 K 0 5mv 2f 2 N cos 0 5 4kg v 2f v f cos m s b v0 1m s K 0 5mv 2f 0 5 4kg 1m s 2 c v0 1m s K 0 5mv 2f 2 J 2 N cos 0 5 4kg v 2f 2 J 2 N cos 0 5 4kg v 2f 2 J v f 1 cos m s v f 1 cos m s 18 In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kg psychology book as the book slides a distance of d 0 5m up a frictionless ramp a During the displacement what is the net work done on the book by Fa the gravitational force on the book and the normal force on the book b If the book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacement y x N d W 0 Only Fgx Fax do …
View Full Document
Unlocking...