Question

A force of 18 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length

Answer 1

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:

$F=kx\\Where, k\to spring\ constant$

Express in terms of 'k'. This gives,

$k=\frac{F}{x}$

Now, plug in 'F' and 'x₁' values and solve for 'k'. This gives,

$k=\frac{18\ lb}{4\ in}=4.5\ lb/in$

Now, work done in stretching or compressing a spring by a length of 'x' is given as:

$Work=\frac{1}{2}kx^2$

Here, $x=x_2=6\ in,k=4.5\ lb/in$. Solve for work, 'W'. This gives,

$W=\frac{1}{2}\times 4.5\ lb/in\times (4\ in)^2\\\\W=\frac{1}{2}\times 4.5\times 16\ \textrm{lb-in}\\\\W=36\ \textrm{lb-in}$

Therefore, the work done in stretching it from its natural length to 6 inch beyond its natural length is 36 lb-in