Final answer:
Using Hooke's Law and conservation of mechanical energy, the speed of a 0.26-kg block can be calculated at different points during its motion in relation to a spring's equilibrium position.
Step-by-step explanation:
When a 0.26-kg block is suspended on a vertical spring and stretches it 3.10 cm, this situation involves simple harmonic motion and the conservation of mechanical energy.
The block's initial and final positions relative to the spring's equilibrium position allow us to calculate its speed at different points in its motion.
We can use Hooke's Law, F = -kx, to find the spring constant k, as F here is the weight of the block (mg), and x is the stretch caused by it (0.031 m). Using that k and the conservation of energy:
E_initial = E_final, where the mechanical energy E is the sum of the potential energy in the spring (1/2)kx^2 and the kinetic energy of the block (1/2)mv^2.
First, the potential energy in the spring when the block stretches the spring 3.10 cm (0.031 m) can be used to find the spring constant k.
Then, the potential energy at the new stretched position (8.10 cm or 0.081 m below equilibrium) is set equal to the sum of the potential energy when the block is 3.60 cm (or 0.036 m) below equilibrium and the kinetic energy at that point to solve for the block's speed.