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Work Lesson 7.5

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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(x i )  x a b xx

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Page 1: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

Work

Lesson 7.5

Page 2: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 3: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 4: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 5: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 6: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 7: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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)

Page 8: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 9: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 10: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 11: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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

Page 12: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

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?

Page 13: Work Lesson 7.5. Work Definition The product of  The force exerted on an object  The distance the object is moved by the force When a force of 50 lbs

Assignment A

• Lesson 7.5• Page 405• Exercises 1 – 41 EOO