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phy2048-ch6_new

# phy2048-ch6_new - Chapter 6 Kinetic energy and work I...

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1 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 I. Kinetic energy Energy associated with the state of motion of an object. ) 1 . 7 ( 2 1 2 mv K = Units: 1 Joule = 1J = 1 kgm 2 /s 2 = N m II. Work Energy transferred “to” or “from” an object by means of a force acting on the object. To barb2right +W From barb2right -W - Constant force: x x ma F = d v v a d a v v x x 2 2 2 0 2 2 0 2 - = + = Work done by the force = Energy transfer due to the force. Energy: scalar quantity associated with a state (or condition) of one or more objects. ) ( 2 1 1 ) ( 2 1 2 0 2 2 0 2 v v m d ma d v v m ma F x x x - = - = = d F W d F K K v v m x x i f = = - = - ) ( 2 1 2 0 2

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2 - 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. ) 3 . 7 ( cos d F d F d F W x arrowrightnosp arrowrightnosp = = = ϕ - Assumptions: 1) F=cte, 2) Object particle-like. 0 90 90 180 90 = - > > + < ringoperator ringoperator ringoperator ringoperator ϕ ϕ ϕ W W Units: 1 Joule = 1J = 1 kgm 2 /s 2 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. 2) F net barb2right W net =F net d Calculation: 1) W net = W 1 +W 2 +W 3 +… F d cos φ II. Work-Kinetic Energy Theorem ) 4 . 7 ( W K K K i f = - = Δ Change in the kinetic energy of the particle = Net work done on the particle III. Work done by a constant force - Gravitational force: ) 5 . 7 ( cos ϕ mgd d F W = = arrowrightnosp arrowrightnosp Rising object : W= mgd cos180º = -mgd barb2right F g transfers mgd energy from the object’s kinetic energy. Falling object: W= mgd cos 0º = +mgd barb2right F g transfers mgd energy to the object’s kinetic energy.
3 IV. Work done by a variable force - External applied force + Gravitational force: ) 6 . 7 ( g a i f W W K K K + = - = Δ 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.

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phy2048-ch6_new - Chapter 6 Kinetic energy and work I...

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