When a force is apply to Spring it got stretched by 25mm. Calculate the work done by F=20A to stretched the spring by 1mm is

The initial stretch $x_1$ caused by the force $F$ is $85 \text{ mm}$.
The target stretch $x_2$ for which work done is to be calculated is $1 \text{ mm}$.
We know that $1 \text{ mm} = 10^{-3} \text{ m}$.

$$x_1 = 85 \text{ mm} = 85 \times 10^{-3} \text{ m} = 0.085 \text{ m}$$ $$x_2 = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} = 0.001 \text{ m}$$

The applied force is $F = 20 \text{ N}$.

Hooke's Law states that the force $F$ required to stretch or compress a spring by a distance $x$ is given by $F = kx$, where $k$ is the spring constant.
Using the given force $F = 20 \text{ N}$ and the corresponding stretch $x_1 = 0.085 \text{ m}$:

$$k = \frac{F}{x_1}$$ $$k = \frac{20 \text{ N}}{0.085 \text{ m}}$$ $$k \approx 235.294 \text{ N/m}$$

The work done to stretch a spring by a distance $x$ from its equilibrium position is given by the formula $W = \frac{1}{2}kx^2$.
We need to find the work done for a stretch of $x_2 = 0.001 \text{ m}$.

$$W = \frac{1}{2} k x_2^2$$ $$W = \frac{1}{2} (235.294 \text{ N/m}) (0.001 \text{ m})^2$$ $$W = \frac{1}{2} (235.294) (0.000001)$$ $$W = 117.647 \times 0.000001$$ $$W = 0.000117647 \text{ J}$$

Rounding to three significant figures:

$$W \approx 1.18 \times 10^{-4} \text{ J}$$

The work done to stretch the spring by $1 \text{ mm}$ is $\boxed{\text{}1.18 \times 10^{-4} \text{ J}}$.