If the work required to stretch a spring 3 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 9 in. beyond its natural length?

3/8

Step-by-step explanation:

Given that:

Work required to stretch a string:

3 ft = 6ft - lb

The work required to stretch it 9 inches

Given that :

Work(W) = kx

Take the integral of W at 3 fts beyond its natural length;

W = ∫kx dx at 0 to 3

W = k ∫xdx at 0 to 3

W = 6

W = k[x²/2] at x =0 to x = 3

6 = k[3² /2] - 0

6 = k[9/2]

12 = 9k

k = 12/9 = 4/3 = 1.333

Converting inches to feet:

1 inch 0.0833 ft

9 inches = 0.75 ft

W = ∫kx dx at 0.75 to 0

W = 4/3 ∫xdx at 0 to 0.75 = 3/4

W = 4/3[x²/2] at x =0.75 to 0

W = 4/3[(3/4)²/2] at 0.75 to 0

W = 4/3[(9/16) / 2]

W = 4/3 * (9/16 * 1/2)

W = 36/96

W = 6/16 = 3/8 ft - lbs