To return a special spring to its equilibrium position A force in the form of F = –(5.0 N/m) X - (15 N/m3) X3 is applied. Here x gives the distance from the equilibrium position. Find the work done by stretching the spring from 0.10 m to 0.20 m.

Q: The students is given 130 cm long ideal spring whose unstretched position is at x=0.00 cm. A student…

A: Spring constant, k=41.1 N/m xi=22.2 cm = 0.222 m xf=30.5 cm =0.305 m

Q: F, (N) 3 2 1 x (m) 8 10 12 14 16 2 4 6 FIGURE P5.60

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