Answer: 240 inch-lbs.
Step-by-step explanation:
The spring constant, k, is given by:
k = (F/x)
where F is the force, x is the displacement, and k is the spring constant.
Using the given information, we can calculate the spring constant as:
k = (80 lbs / 2 in) = 40 lbs/in
To find the work done in stretching the spring from its relaxed state to a total of 4 inches, we need to integrate the force over the displacement:
W = ∫F dx
Since the force required to stretch the spring varies with displacement, we need to break up the integration into two parts: one from 0 to 2 inches, and another from 2 to 4 inches.
W = ∫0^2 F dx + ∫2^4 F dx
The first integral is:
∫0^2 F dx = ∫0^2 (kx) dx = (1/2)kx^2 = (1/2)(40 lbs/in)(2 in)^2 = 80 inch-lbs
The second integral is:
∫2^4 F dx = ∫2^4 (2k) dx = 2kx = 2(40 lbs/in)(4 in - 2 in) = 160 inch-lbs
Therefore, the total work done in stretching the spring from its relaxed state to a total of 4 inches is:
W = ∫0^2 F dx + ∫2^4 F dx = 80 inch-lbs + 160 inch-lbs = 240 inch-lbs
So, the work done in stretching the spring from its relaxed state a total of 4 inches is 240 inch-lbs.