Tutorial Exercise A force of 9 lb is required to hold a spring stretched 3 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length? Step 1 According to Hooke's Law, the force required to maintain a spring stretched x units beyond its natural length is proportional to x. This means that f(x) = kx, for some constant k. ¹k. So Since our spring is stretched 3 in., which is equal to 1/4 ft, we have 9 = 4 k = 3 x lb/ft. Submit Skip (you cannot come back)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 67E
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Tutorial Exercise
A force of 9 lb is required to hold a spring stretched 3 in. beyond its natural length. How much work W is done in
stretching it from its natural length to 5 in. beyond its natural length?
Step 1
According to Hooke's Law, the force required to maintain a spring stretched x units beyond its natural length is
proportional to x. This means that f(x) = kx, for some constant k.
Since our spring is stretched 3 in., which is equal to 1/4
ft, we have 9 = =k. So
4
k = 3
x lb/ft.
Submit Skip (you cannot come back)
Transcribed Image Text:Tutorial Exercise A force of 9 lb is required to hold a spring stretched 3 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length? Step 1 According to Hooke's Law, the force required to maintain a spring stretched x units beyond its natural length is proportional to x. This means that f(x) = kx, for some constant k. Since our spring is stretched 3 in., which is equal to 1/4 ft, we have 9 = =k. So 4 k = 3 x lb/ft. Submit Skip (you cannot come back)
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