Work Energy and Power

Spring constant, $k = 5 \times 10^3 \mathrm{N/m}$Initially stretched position, $x_1 = 10 \mathrm{cm} = 0.1 \mathrm{m}$Further stretched position, $x_2 = 20 \mathrm{cm} = 0.2 \mathrm{m}$Work Done:The work done to further stretch the spring from $x_1$ to $x_2$ is equal to the change in potential energy of the spring. This can be calculated using the formula:$\; \text{ Work done } \; = \; \frac{1}{2} k \left[ (x_2)^2 - (x_1)^2 \right]$Substitute the given values:$\; = \; \frac{1}{2} \times 5 \times 10^3 \left[ (0.2)^2 - (0.1)^2 \right]$$\; = \; \frac{1}{2} \times 5 \times 10^3 \times (0.04-0.01)$$\; = \; \frac{1}{2} \times 5 \times 10^3 \times 0.03$$\; = 2.5 \times 10^3 \times 0.03$$\; = 75 \, \mathrm{N \cdot m}$Therefore, the work required to stretch the spring further by another 10 cm is $75 \, \mathrm{N \cdot m}$.