Answer:
the work done in stretching the spring from its natural length is -1.8 ft.lb
Step-by-step explanation:
Given the data in the question;
From Hook's law, the magnitude of force applied on a spring is expressed as follows;
|F| = k|x|
where F is the force applied the spring, k is the spring constant, x is the length extension
Now, given that; F = 4 pounds = 4 lb and x = 0.4 ft
we substitute
4 = k × 0.4
k = 4 / 0.4
k = 10 lb/ft
Also, Hook's law can be expressed as follows;
F = -kx ---------- let this be equation 1
Now, work done by a force applied between two points is expressed as follows;
dW = Fdx ----------- let this be equation 2
substitute equation 1 into equation 2
dW = (-kx )dx
we integrate both side
∫(dW) = ∫((-kx )dx)
W = -k∫xdx
W = -k[tex]\frac{x^2}{2}[/tex]
w = -[tex]\frac{1}{2}[/tex]kx²
we substitute 10 lb/ft for k and 0.6 ft for x
w = -[tex]\frac{1}{2}[/tex] × 10 lb/ft × (0.6 ft)²
w = -[tex]\frac{1}{2}[/tex] × 10 lb/ft × 0.36 ft²
w = -1.8 ft.lb
Therefore, the work done in stretching the spring from its natural length is -1.8 ft.lb
