A spring stretches 0.05 m under a force of 40 N. Find the energy stored in it. A spring has a Force constant k = 500 N/m. Find the energy stored in the spring when it is stretched by 0.02 m.

The formula for Hooke's Law is $F = kx$, where $F$ is the force, $k$ is the spring constant, and $x$ is the extension.

$$k = \frac{F}{x}$$

Substitute the given values:

$$k = \frac{40 \, \text{N}}{0.05 \, \text{m}}$$ $$k = 800 \, \text{N/m}$$

The formula for energy stored in a spring is $U = \frac{1}{2}kx^2$.
Substitute the calculated spring constant and the given extension:

$$U = \frac{1}{2} (800 \, \text{N/m}) (0.05 \, \text{m})^2$$ $$U = \frac{1}{2} (800 \, \text{N/m}) (0.0025 \, \text{m}^2)$$ $$U = 400 \times 0.0025 \, \text{J}$$ $$U = 1 \, \text{J}$$

The energy stored in the spring is $\boxed{\text{1 J}}$.

** (Ex 2):**
A spring has a Force constant $k = 500 \, \text{N/m}$. Find the energy stored in the spring when it is stretched by $0.02 \, \text{m}$.

The formula for energy stored in a spring is $U = \frac{1}{2}kx^2$.
Substitute the given values:

$$U = \frac{1}{2} (500 \, \text{N/m}) (0.02 \, \text{m})^2$$ $$U = \frac{1}{2} (500 \, \text{N/m}) (0.0004 \, \text{m}^2)$$ $$U = 250 \times 0.0004 \, \text{J}$$ $$U = 0.1 \, \text{J}$$

The energy stored in the spring is $\boxed{\text{0.1 J}}$.