Post on 18-Jan-2018
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Work
Lesson 7.5
Work
• DefinitionThe product of The force exerted on an object The 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 done
50
12 ft
Hooke's Law
• Consider the work done to stretch a spring
• Force required is proportional to distance When k is constant of proportionality Force to move dist x = k • x = F(x)
• Force required to move through i th
interval, x W = F(xi) x
a b
x
Hooke's Law
• We sum those values using the definite integral
• The work done by a continuous force F(x) Directed along the x-axis From x = a to x = b
( )b
a
W F x dx
Hooke's Law
• A spring is stretched 15 cm by a force of 4.5 N How much work is needed to stretch the spring
50 cm?• What is F(x) the force function?
• Work done?
4.5 0.1530( ) 30
F k xkk
F x x
0.5
0
30W xdx
Winding Cable
• Consider a cable being wound up by a winch Cable is 50 ft long 2 lb/ft How much work to wind in 20 ft?
• Think about winding in y amt y units from the top 50 – y ft hanging dist = y force required (weight) =2(50 – y)
20
0
2 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 liquid Determine weight Determine dist moved Take 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 top
a
br
2
0
( )a
W r b y dy
Pumping Liquids
• Suppose tank has r = 4 height = 8 filled with petroleum (54.8 lb/ft3)
• What is work done to pump oil over top Disk weight? Distance moved? Integral?
84
54.8 16Weight y (8 – y)8
0
54.8 16 (8 )Work y y
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 is
2W F x p r x
Work 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 and
1
0
V
V
W p dV
1
0
V
V
kW 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 EOO