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Mathematics 27 Online
OpenStudy (anonymous):

need an explanation: suppose 2J of work is needed to stretch spring w/ natural length of 30 cm to 42 cm. how much work is needed to stretch spring from 35cm to 40 cm? how far beyond natural length will force of 30 N keep spring stretched?

OpenStudy (anonymous):

work done on spring = 0.5 kx^2 = 2J, where k is the spring constant and x is the extension from rest thus find the spring constant of the spring...as it implies it is a constant: thus 2 = 0.5 kx^2 = 0.5 *k*0.12^2 (extension from rest = (42 -30)cm = 12cm =0.12m so 1 = k*0.12^2 and k = 1/0.12^2 = 69.4 N/m Now you can find the work for an extension of 35cm to 40 cm(0.35m to 0.40 m) work = 0.5 69.4(0.05)^2 = 0.08675 J of work how far beyond natural length for a force of 30 N: Now F=kx, k = 69.4, therefore x = F/k = 30/69.4 = 0.432m = 43.2cm

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