force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 11 in. beyond its natural length?

The force required to stretch a spring is given by Hooke's Law, which states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is:

F = kX

where:

• F is the force applied,
• X is the distance the spring is stretched or compressed, and
• k is the spring constant.

From the problem, we know that a force of 8 lb is required to hold the spring stretched 8 in. beyond its natural length. We can use this information to find the spring constant k:

8 lb = k * 8 in k = 8 lb / 8 in = 1 lb/in

Now, we want to find the work done in stretching the spring from its natural length to 11 in. beyond its natural length. The work done on an object is given by the integral of the force over the distance over which it is applied. In this case, the force is not constant, but varies with the distance the spring is stretched. Therefore, we need to integrate the force from 0 to 11 in:

W = ∫ from 0 to 11 of F dx = ∫ from 0 to 11 of kX dx = ∫ from 0 to 11 of (1 lb/in * X) dx = [0.5 * X^2] from 0 to 11 = 0.5 * (11 in)^2 - 0.5 * (0 in)^2 = 0.5 * 121 in^2 = 60.5 in-lb

So, the work done in stretching the spring from its natural length to 11 in. beyond its natural length is 60.5 in-lb.