WorkSlide 2Hooke's LawSlide 4Slide 5Winding CablePumping LiquidsPumping Liquids – GuidelinesPumping LiquidsWork Done by Expanding GasSlide 11Slide 12Assignment AWorkLesson 7.5Work•DefinitionThe product ofThe force exerted on an objectThe distance the object is moved by the force•When a force of 50 lbs is exerted to move an object 12 ft.600 ft. lbs. of work is done5012 ftHooke's Law•Consider the work done to stretch a spring•Force required is proportional to distanceWhen k is constant of proportionalityForce to move dist x = k • x = F(x)•Force required to move through i thinterval, xW = F(xi) xabxHooke's Law•We sum those values using the definite integral•The work done by a continuous force F(x)Directed along the x-axisFrom x = a to x = b( )baW F x dx=�Hooke's Law•A spring is stretched 15 cm by a force of 4.5 NHow much work is needed to stretch the spring 50 cm?•What is F(x) the force function?•Work done?4.5 0.1530( ) 30F k xkkF x x= �= �==0.5030W x dx=�Winding Cable•Consider a cable being wound up by a winchCable is 50 ft long2 lb/ftHow much work to wind in 20 ft?•Think about winding in y amty units from the top 50 – y ft hanging dist = y force required (weight) =2(50 – y)( )2002 50W y dy= -�Pumping Liquids•Consider the work needed to pump a liquid into or out of a tank•Basic concept: Work = weight x dist moved•For each V of liquidDetermine weightDetermine dist movedTake summation (integral)Pumping Liquids – Guidelines•Draw a picture with thecoordinate system•Determine mass of thinhorizontal slab of liquid•Find expression for work needed to lift this slab to its destination•Integrate expression from bottom of liquid to the topabr20( )aW r b y dyr p= �� -�Pumping Liquids •Suppose tank hasr = 4height = 8filled with petroleum (54.8 lb/ft3)•What is work done to pump oil over topDisk weight?Distance moved?Integral?8454.8 16Weight yp= �� �D(8 – y)8054.8 16 (8 )Work y yp= �� � - D�Work Done by Expanding Gas•Consider a piston of radius r in a cylindrical casing as shown here•Let p = pressure in lbs/ft2•Let V = volume of gas in ft3•Then the work incrementinvolved in moving the pistonΔx feet is2W F x p r xpD = �D = � �DWork Done by Expanding Gas•So the total work done is the summation of all those increments as the gas expands from V0 to V1•Pressure is inversely proportionalto volume so p = k/V and10VVW p dV= ��10VVkW dVV= ��Work Done by Expanding Gas•A quantity of gas with initial volume of1 cubic foot and a pressure of 2500 lbs/ft2 expands to a volume of 3 cubit feet.•How much work was done?Assignment A•Lesson 7.5•Page 405•Exercises 1 – 41
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