Final answer:
To calculate the work needed to stretch a spring 6 inches beyond its equilibrium, we use the fact that the work required is proportional to the square of the extension. Since the work for a 12-inch extension is 10 ft-lb, the work for a 6-inch extension is a quarter of that, resulting in 2.5 ft-lb.
Step-by-step explanation:
The student's question is related to the work done to stretch a spring in the field of physics. When a force is applied to stretch a spring, the work done on the spring is quantified by the equation W = ½kx², where W is the work, k is the spring constant, and x is the displacement of the spring from its equilibrium length.
Since it is given that the work required to stretch a spring 1 ft (12 inches) beyond its natural length is 10 ft-lb, we can use this information to find the work needed to stretch it half of that distance (6 inches). According to the properties of springs, the work needed to stretch it further is directly proportional to the square of the extension. Therefore, to find the work W needed to stretch the spring 6 inches, we can apply the relationship:
W required to stretch from 0 to 12 inches is 10 ft-lb
W required to stretch from 0 to 6 inches will be ¼ of the work done to stretch from 0 to 12 inches because (6/12)² = ¼
So, W needed to stretch the spring 6 inches is:
W = (1/4) × 10 ft-lb = 2.5 ft-lb