Hooke’s law states that when a spring is stretched, it pulls back with a force proportional to the amount of the stretch. The constant of proportionality is a characteristic of the spring known as the spring constant. Thus a spring with spring constant k exerts a force f(x2)= kx when it is stretched a distance x. A certain spring has spring constant k =20 lb/ft. Find the work done when the spring is pulled so that the amount by which it is stretched increases from x =0 to x = 2 ft.

Hooke’s law states that when a spring is stretched, it pulls back with a force proportional to the amount of the stretch. The constant of proportionality is a characteristic of the spring known as the spring constant. Thus a spring with spring constant k exerts a force f(x2)= kx when it is stretched a distance x. A certain spring has spring constant k =20 lb/ft. Find the work done when the spring is pulled so that the amount by which it is stretched increases from x =0 to x = 2 ft.

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Question

A certain spring has spring constant k =20 lb/ft. Find the work done when the spring is pulled so that the amount by which it is stretched increases from x =0 to x = 2 ft.

Expert Solution

Step 1

We need to find the amount of work done in pulling the spring from x=0 to x=2 if the spring with spring constant k exerts a force $f (x) = k x$ when it is stretched a distance x .

The work done in moving an object from a to b is defined as the limit of this quantity as $n \to \infty$ .

$w = \lim_{n \to \infty} \sum_{m = 1}^{n} f (x_{n}) ∆ x$

$∆ x = \frac{b - a}{n}$ and $x_{m} = a + m ∆ x$

To determine the work done, we first need to evaluate the value of $∆ x$ and $x_{k}$ .

We have $∆ x = \frac{b - a}{n}$ .

On substituting $a = 0, b = 2$ , we get

$\begin{array}{rcl} ∆ x & = & \frac{2 - 0}{n} \\ = & \frac{2}{n} \end{array}$

On substituting a=0 and $∆ x = \frac{2}{n}$ in $x_{m}$ , we get

$\begin{array}{rcl} x_{m} & = & 0 + m (\frac{2}{n}) \\ = & \frac{2 m}{n} \end{array}$

Step 2

On substituting $\begin{array}{rcl} x_{m} & = & \frac{2 m}{n} \end{array}$ in $f (x_{m})$ we get

$\begin{array}{rcl} f (x_{m}) & = & k x_{m} \\ = & k \times \frac{2 m}{n} \\ = & \frac{2 k}{n} . m \end{array}$

On substituting the value of $f (x_{m})$ in $w = \lim_{n \to \infty} \sum_{m = 1}^{n} f (x_{m}) ∆ x$ , we get

$\begin{array}{rcl} w & = & \lim_{n \to \infty} \sum_{m = 1}^{n} f (x_{n}) ∆ x \\ = & \lim_{n \to \infty} \sum_{m = 1}^{n} \frac{2 k}{n} . m . \frac{2}{n} \\ = & \lim_{n \to \infty} \sum_{m = 1}^{n} \frac{4 k}{n^{2}} . m \\ = & \lim_{n \to \infty} \frac{4 k}{n^{2}} \sum_{m = 1}^{n} m \end{array}$

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