Lesson 22 Work Done by A Apring and Pump2 Revised (1)

# Lesson 22 Work Done by A Apring and Pump2 Revised (1) - I...

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Click to edit Master subtitle style Integral Cal culus n Work done by a spring n Work done by pumping a liquid

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Work done by a spring
Hooke’s Law Hooke ’s Law states that within the limits of elasticity the displacement produced in a body is proportional to the force applied, that is, F = kx, where the constant k is the constant of proportionality called the modulus. Thus , F = kx The work done is

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1. I f the modulus of a spring is 20 lbs./ in., what is the work required to stretch the spring a distance of 6 inches? F(x) = 20x W = 6 20xdx = ½ 20 x ] = 10(6)2 - 10(0)2 = 10 (36) W = 360 lbs. in. 0 6 0
2. I f a force of 50 lbs. stretches a 12 in. spring to 14 in., find the work done in stretching the spring from 15 in. to 17 in. To get the limits of integration 15-12 = 3 17-12 = 5 12 in 15in 17in if x = 2, F = 50 F = kx W = 5 25xdx 50 = 2k = ½ 25x2 ]35 K = 25 = 25(5)2 /2 - 25(3) 2 /2 = 625/2 - 225/2 3 x = 0 x = 3 x = 5 F(x)= 25x

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3. A spring has a natural length of 10 inches. An 800-lb force stretches the spring 14-inches. (a) Find the force constant. (b)
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Lesson 22 Work Done by A Apring and Pump2 Revised (1) - I...

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