Question

A force of 10 lb is required to hold a spring stretched 1/3ft beyond its natural length. Which of the following integrals represents the amount of work done in stretching it from its natural length to 1/2 ft beyond its natural length? 5/6 30x da J1/2 -1/2 30x da 0 so" [" -1/2 20. dac 1/2 10 zda 0 3 Com S 5/6 203 do 1/2

Answer 1

solution - Force Required to stretch spring is ginen hook's by law :- f(x) = ka to stretched 43 ft beyond its hatural length force Required f(x) = 10 lb f(a) k eb R al- ft K 30 9ь so Work donc in stretching from its natural length to 1₂ ft beyond its natural length is given by x2 s f(x) da R = ft So f(x)= W= 3ox dn R