Question

A vertical spring of force constant $100\mathrm{N}/\mathrm{m}$ is attached to a hanging mass of $10\mathrm{kg}$. Now an external force is applied on mass so that the spring is stretched (maximum) by additional $2\mathrm{m}$. The work by the force F is ($\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$)

Answer 1

Step 1: Given information

Force constant, $\mathrm{k}=100\mathrm{N}/\mathrm{m}$

Mass, $\mathrm{m}=10\mathrm{kg}$

Extension in the length of spring, $\mathrm{x}=2\mathrm{m}$

$\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$

Step 2: To find

We have to determine the work by the force.

Step 3: Calculation

We know the formula,

Work done,

$\Rightarrow \mathrm{W}=\frac{1}{2}{\mathrm{kx}}^2$

Where,

F is the spring force,

x is the displacement from the equilibrium position.

k is the spring constant.

The spring constant is the characteristic property of the spring. It depends upon the material of construction and is measured in the units of $\mathrm{N}{\mathrm{m}}^{-1}$.

By substituting the given values in the above formula, we get

$$\begin{gathered} =\frac{1}{2}\times 100\times {\left(2\right)}^2 \\ =\frac{1}{2}\times 100\times 4 \\ =100\times 2 \\ =200\mathrm{J} \end{gathered}$$

Therefore, the work done by the force is $200\mathrm{J}$.