# A force of 8 lb is required to hold a spring stretched 8 inches beyond its natural length. How...

## Question:

A force of 8 lb is required to hold a spring stretched 8 inches beyond its natural length. How much work is done in stretching it from its natural length to 10 inches beyond its natural length?

## Work:

Work is the amount of energy it takes to move an object from one place to another. We compute it using a line integral:

{eq}W = \int_C \vec F \cdot d\vec r {/eq}

For linear displacement, this simplifies to

{eq}\begin{align*} W = \int_a^b F\ dx \end{align*} {/eq}

Recall Hooke's Law which says that for springs, the force is proportional to the displacement, i.e.

{eq}\begin{align*} F = kx \end{align*} {/eq}

We know that {eq}F = 8 {/eq} lbs when {eq}x=8 {/eq} in. Inches are not very standard, so we'll use the equivalent {eq}x = \frac23 {/eq} ft instead. Now we can use Hooke's Law to find the general force equation at play here.

{eq}\begin{align*} F &= kx \\ 8 &= k \cdot \frac23 \\ k &= 12 \end{align*} {/eq}

so

{eq}\begin{align*} F = 12x \end{align*} {/eq}

Then the work it takes to stretch it from equilibrium ({eq}x=0 {/eq}) to 10 inches, i.e. {eq}x = \frac56 {/eq} ft. is

{eq}\begin{align*} W &= \int_0^{5/6} 12x\ dx \\ &= \left [ 6x^2 \right ]_0^{5/6} \\ &= 6 \left( \frac56 \right)^2 \\ &= \frac{25}{6} \\ &\approx 4.167\ \text{ft} \cdot \text{lbs} \end{align*} {/eq}