Question

The work done in joules in increasing the extension of a spring of stiffness $10N/cm$ from $4cm$ to $6cm$.

Answer 1

Step 1: Given data

Spring constant = $k=10N/cm\Rightarrow 10\times 100N/m\because \left[1m=100cm\right]$

$x_2=6cm=\frac{6}{100}m$ is the final position, $x_1=4cm=\frac{4}{100}m$ is the initial position

Step 2: Use the formula of work done by a spring

Work done by a spring$=\frac{1}{2}k(x_2^2-x_1^2)$ where,

$k$ is stiffness of spring, $x_1$is the initial position, $x_2$ is the final position

Step 3: Solution

Work done by a spring

$=\frac{1}{2}\times 1000\times \left[{\left(0.06\right)}^2-{\left(0.04\right)}^2\right]$

$$\begin{gathered} =\frac{1}{2}\times 1000\times \left[0.0036-0.0016\right] \\ =\frac{1}{2}\times 1000\times 0.0020 \\ =1J \end{gathered}$$

Hence, work done by the spring is $1$ joule.