Answer:
W = 36 lb-in
Explanation:
Given:
Force required to pull a spring (F) = 18 lb
Elongation in the spring (x₁) = 4 in
New elongation (x₂) = 6 in
We know that, the force required to stretch or compress the spring is given by the formula:
[tex]F=kx\\Where, k\to spring\ constant[/tex]
Express in terms of 'k'. This gives,
[tex]k=\frac{F}{x}[/tex]
Now, plug in 'F' and 'x₁' values and solve for 'k'. This gives,
[tex]k=\frac{18\ lb}{4\ in}=4.5\ lb/in[/tex]
Now, work done in stretching or compressing a spring by a length of 'x' is given as:
[tex]Work=\frac{1}{2}kx^2[/tex]
Here, [tex]x=x_2=6\ in,k=4.5\ lb/in[/tex]. Solve for work, 'W'. This gives,
[tex]W=\frac{1}{2}\times 4.5\ lb/in\times (4\ in)^2\\\\W=\frac{1}{2}\times 4.5\times 16\ \textrm{lb-in}\\\\W=36\ \textrm{lb-in}[/tex]
Therefore, the work done in stretching it from its natural length to 6 inch beyond its natural length is 36 lb-in
