How much work needed to stretch is 9in beyond its natural length?

Question

If the work required to stretch a spring 1ft beyond its natural length is 12gt-lb, how much work is needed to stretch it 9in beyond its natural length?

Answer 1

The answer is 6.75 gt-lb,

The work is proportional to the square of the extension ' x.' That is because the force F is itself proportional to x (k x, say) , and the work done is the integral of k x dx, that is [1/2 k x^2].

So it takes (9/12)^2 x 12gt-lb = 9/16 x 12gt-lb = 27 / 4 or 6.75 gt-lb.

Live long and prosper.

Answer 2

F = kx
W = ∫Fdx
x
∫kxdx = (k/2)x^2
0
W = (k/2)9^2
12 = (k/2)12^2
W/12 = 81/144
W = 81/12 = 27/4 = 6.75 gt-lb
(whatever a gt-lb is . . .)

Answer 3

you take advantage of proportions a million feet. / 15 feet*lb = .5 feet/ x feet*lb --> a million/15 lb = .5/ x lb. x = 7.5 lb yet you could also seem on the area stretched and draw your conclusions using deductive reasoning.... 15 lb*feet for a million feet so 7.5 lb*feet for .5 feet