Answer:
120 pounds
Explanation:
Data provided in the question:
Force applied, F₁ = 15 pounds
Displacement by Force F₁, x₁ = 1 foot
At natural length of the spring total movement of the door = 8 feet
Now,
we know,
Force in spring, F = kx
where, k is the spring constant
Thus,
15 = k(1) ............(1)
now when the door travels half the distance x₂ =
= 4 feet
Therefore,
F₂ = k(4) ..........(2)
Dividing 2 by 1
we get

or

or
F = 15 × 4 = 60 pounds
Thus,
for 2 springs combined force applied = 2 × 60 pounds = 120 pounds