PHY11T1WORK - WORK, ENERGY & POWER WORK (W) (Mechanical)...

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WORK, ENERGY & POWER

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WORK (W) (Mechanical) Work – the amount of energy transferred by a force acting through a distance (displacement) – the scalar product of the applied force (F) on an object and the displacement (s) this object has moved to. F s F W = Fscosθ θ θ Since Weight also uses “W” we will use the small case w for weight and the big case W for work w = mg
WORK (W) Where  θ  is the angle between  the force & displacement vector F s θ W = F s cos θ Units : MKS – Joule (J) , 1  J = 1 N-m  CGS – Ergs , 1  Erg = 1 dyne-cm  NOTE : No work is done if the force is perpendicular  to the displacement  (  θ  is  90  or  270 )

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WORK (W)  Work due to several forces = Algebraic sum of all work or Net Work W TOTAL  = W Net s m F w  = mg N f W net  = F net  s For this example  F net  = Fcos θ  – f and f = µ(mg-Fsin θ ) W net  = mas F net  = ma a
WORK (W)  Example : A block of 20 kg is pulled over a distance of 10m at an  acceleration of 2 m/s 2 ,using a constant pulling force  F  of 200N that  makes an angle of 22° with the horizontal. (a)What is total work done?  (b)What is the opposite work done by friction? S = 10 m m F = 200  N w  = mg N f   = 22° S = 10 m w  = mg N f F = 200  N a = 2 m/s 2

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WORK (W)  (a) W T = mas = (20 kg)(2 m/s 2 )(10m) = 400 Joules = 400 J (b) W f = fs cosθ f , θ f = 180°  F net  = Fcos θ  - f   (see “friction notes – object push or pulled by a force at an angle”)        F net  = ma   (see “Newton’s Second Law “)  F net  = (20 kg)(2 m/s 2 ) = 40 N  f = Fcos θ  − F net  = (200N)(cos 22°) − 40 N  f = Fcos θ  − F net  f = 145.4 N W f  = fs cos 180° = (145.4N)(10m)(-1) f S = 10 m W f  =  −1,454 J
ENERGY   From the greek word  energeia  – meaning “activity” or “operation”

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This note was uploaded on 05/19/2011 for the course PHYSICS 10 taught by Professor Darp during the Spring '11 term at Mapúa Institute of Technology.

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PHY11T1WORK - WORK, ENERGY & POWER WORK (W) (Mechanical)...

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