Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
[tex]W =k\int_{6}^{10.1} x[/tex]
[tex]W = \dfrac{k}{2}[x^2]_6^{10.1}[/tex]
[tex]W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)[/tex]
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.