A 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?W = ft-lb

The problem involves Hooke's Law, which states that the force required to stretch or compress a spring by some distance is proportional to that distance. That is, F = kx, where F is the force, k is the spring constant, and x is the distance.

Step 1: Find the spring constant (k) Given that a force of 8 lb is required to hold the spring stretched 8 in. beyond its natural length, we can use Hooke's Law to find the spring constant.

8 lb = k * 8 in.

Solving for k, we get k = 1 lb/in.

Step 2: Calculate the work done (W) The work done on a spring is given by the formula W = 1/2 * k * x^2, where x is the distance the spring is stretched from its natural length.

In this case, we want to find the work done in stretching the spring from its natural length to 11 in. beyond its natural length.

So, W = 1/2 * 1 lb/in * (11 in)^2 = 60.5 lb-in.

However, the question asks for the answer in foot-pounds (ft-lb). Since 1 ft = 12 in, we need to convert our answer to ft-lb.

W = 60.5 lb-in * (1 ft/12 in) = 5.04 ft-lb.

So, the work done in stretching the spring from its natural length to 11 in. beyond its natural length is approximately 5.04 ft-lb.