Friday, May 18, 2018

Single Variable Calculus, Chapter 6, 6.4, Section 6.4, Problem 10

If 12ftlb is the required work to stretch a spring 1 ft beyond its natural length, how much work is needed to sketch 9 inches beyond its vertical length?

Recall from Hooke's Law,

f(x)=kx where x is the maximum elongated length of the spring and k is the spring constant


W=baf(x)dxW=bakxdxW=10kxdx; recall that 9 inches (1ft12inches)12=k[x22]1012=k([(1)22][(0)22])12=k(12)k=24lbft


Therefore, the required work to stretch the string to 9 inches is...


W=340kxdx; recall that 9 inches (1ft12inches)=34ftW=34024xdxW=274ftlb

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