Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
11.47J
Hooke's law states that the force or load (F) applied to an elastic material (e.g a spring) is directly proportional to the extension or compression (e) caused by the load. i.e
F ∝ e
F = k e ---------------(i)
where;
k = proportionality constant called the spring or elastic constant.
From the question,
F = Force = 10 lb
[Remember that 1 lb = 4.45 N]
=> F = 10 x 4.45 N = 44.5 N
Also;
e = extension = 4 in
[Remember that 1 in = 0.0254 m]
e = 4 x 0.0254 m = 0.1016 m
Substitute these values into equation (i) to get the spring constant of the spring as follows;
44.5 = k x 0.1016
k = [tex]\frac{44.5}{0.1016}[/tex]
k = 439N/m
Now, lets calculate the work done
The workdone W, in stretching a spring from its natural length to some length beyond the natural length, is given by;
W = [tex]\frac{1}{2}[/tex] x k x e² --------------------(ii)
Where;
k = the spring's constant = 439N/m [as calculated above]
e = extension = 9 in = 9 x 0.0254m = 0.2286m [Recall: 1 in = 0.0254m]
Substitute these values into equation (ii) as follows;
W = [tex]\frac{1}{2}[/tex] x 439 x (0.02286)²
W = 11.47 Nm or 11.47J
Therefore, the work done is 11.47J
