I apologize for the incorrect answer. Let me try to solve the problem again.
The force required to hold a spring stretched x inches beyond its natural length is given by Hooke's law:
F = kx
where F is the force, k is the spring constant, and x is the displacement.
In this case, we are given that a force of 10 lb is required to hold the spring stretched 2 in. beyond its natural length. This means that the spring constant k is:
k = F/x = 10/2 = 5 lb/in.
To find the work done in stretching the spring from its natural length to 6 in. beyond its natural length, we need to integrate the force with respect to displacement:
W = ∫F dx
W = ∫kx dx
W = (1/2)kx^2 + C
where C is the constant of integration.
Evaluating the integral with the limits of integration from 0 to 6, we get:
W = (1/2)k(6^2 - 0^2)
W = 90 in-lb
Therefore, the work done in stretching the spring from its natural length to 6 in. beyond its natural length is 90 in-lb.