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Unformatted text preview: Integral Calculus Work done by a spring Work done by pumping a liquid Work done by a spring Hooke’s Law Hooke’s Law states that within the lim its 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 1. If 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. 6 2. If 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 lim its of integration 1512 = 3 1712 = 5 12 in 15in 17in if x = 2, F = 50 F = kx W = ∫ 5 25xdx 50 = 2k = ½ 25x 2 ] 3 5 K = 25 = 25(5) 2 /2  25(3) 2 /2 = 625/2  225/2 W = 200 lbs.in. 3 x=0 x=3 x=5 F(x)= 25x 3. A spring has a natural length of 10 inches. An 800lb force stretches the spring 14inches. (a) Find the force constant. (b) How much work is done in stretching the spring from 10 inches to 12 inches? (c) How far done in stretching the spring from 10 inches to 12 inches?...
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This note was uploaded on 10/19/2011 for the course MATH 22 taught by Professor Ma'amzapanta during the Fall '11 term at Mapúa Institute of Technology.
 Fall '11
 Ma'amZapanta
 Calculus

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