A spring of spring constant 5 x10³ N /m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

Correct Option: The correct answer is 18.75 N-m.

Step-by-step Explanation:

The work done $W$ in stretching a spring with spring constant $k$ from an initial extension $x_1$ to a final extension $x_2$ is given by the change in its potential energy:

$W=\frac{1}{2}k(x_2^2-x_1^2)$

Let's identify the given values from the problem statement:

1. Spring constant, $k=5\times {10}^3\text{N/m}$

2. Initial extension, $x_1=5\text{cm}=0.05\text{m}$

3. The spring is stretched further by another $5\text{cm}$, so the final extension is:

$x_2=5\text{cm}+5\text{cm}=10\text{cm}=0.10\text{m}$

Now, we substitute these values into the work formula:

$W=\frac{1}{2}\times (5\times {10}^3\text{N/m})\times [{\left(0.10\text{m}\right)}^2-{\left(0.05\text{m}\right)}^2]$

Calculate the squares of the extensions:

${\left(0.10\right)}^2=0.01$

${\left(0.05\right)}^2=0.0025$

Now, find the difference between these squares:

$0.01-0.0025=0.0075$

Substitute this back into the work equation:

$W=\frac{1}{2}\times 5000\times 0.0075$

$W=2500\times 0.0075$

$W=18.75\text{N-m}$

Thus, the work required to stretch the spring by another 5 cm is 18.75 N-m.