A catapult launches a 5 g stone with an average force of 70 N over a distance of 7 cm. Calculate the stone's speed upon release.

The energy stored in the catapult is equal to the work done to stretch it.
Given:
Average force, $F_{avg} = 70 \text{ N}$
Distance stretched, $x = 7 \text{ cm} = 0.07 \text{ m}$

The elastic potential energy ($U$) is:

$$U = F_{avg} \times x$$ $$U = 70 \text{ N} \times 0.07 \text{ m}$$ $$U = 4.9 \text{ J}$$

When the stone is released, the elastic potential energy stored in the catapult is converted into the kinetic energy of the stone.

$$U = K$$

Where $K$ is the kinetic energy of the stone, given by $K = \frac{1}{2} m v^2$.
Given:
Mass of stone, $m = 5 \text{ g} = 0.005 \text{ kg}$

So, we have:

$$4.9 \text{ J} = \frac{1}{2} m v^2$$ $$4.9 = \frac{1}{2} (0.005) v^2$$

$$4.9 = 0.0025 v^2$$ $$v^2 = \frac{4.9}{0.0025}$$ $$v^2 = 1960$$ $$v = \sqrt{1960}$$ $$v \approx 44.27 \text{ m/s}$$

The speed of the stone at the time of release is approximately $44.27 \text{ m/s}$.