If 12ft−lb 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=274ft−lb
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