Question

A spring of spring constant $5\times {10}^3\text{N/m}$ is stretched initially by $5\text{cm}$ from the unstretched position. The work required to further stretch the spring by another $5\text{cm}$ is:

• A. $6.24\text{N-m}$
• B. $12.50\text{N-m}$
• C. $18.75\text{N-m}$
• D. $25.00\text{N-m}$

Answer 1

The correct option is C $18.75\text{N-m}$

Given:
Since, spring's Initial extension is $5\text{cm}$ and further it is extend by $5\text{cm}$.

So, $x_1=5\text{cm};x_2=10\text{cm};k=5\times {10}^3\text{N/m}$

On applying work-energy theorem,

work done by all forces = change in kinetic energy

$\Rightarrow W_{\text{spring}}+W_{\text{external force}}=K.E_f-K.E_i$

$\Rightarrow -\frac{1}{2}k(x_2^2-x_1^2)+W_{\text{external force}}=0-0$

$\Rightarrow -\frac{1}{2}\times 5\times {10}^3\times ({0.1}^2-{0.05}^2)+W_{\text{external force}}=0$

$\Rightarrow W_{\text{external force}}=18.75\text{N}$

Hence, option $\left(c\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Concept - In case of conservative forces (like spring force)}\\ \text{work done against it is stored as potential energy.}\\ \end{array}$