A force of 6 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How muchwork (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its naturallength?XBasic Funcs TrignX↑xA9 USO010197Z23WeSd77CCDC$4rf%5Vt36.06b↓yh&IN7▶un8D(9km<P

F = -kx 6 = -k(0.2) So, k = 30 Work done, W = (1/2)kx2 = (1/2)*(30)*(0.9)2 = 12.15

Answered: A force of 6 pounds is required to hold…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 73E

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A force of 6 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its natural length?