Final answer:
The question involves a 2.10 kg block attached to an ideal spring with force constant of 313 N/m. Given the initial velocity and displacement of the block, we can determine various properties of the system using the equations of motion for a block and a spring.
Step-by-step explanation:
The question involves a 2.10 kg block attached to an ideal spring with force constant of 313 N/m. Given that the block has an initial velocity of -4.00 m/s and a displacement of 0.230 m, we can determine various properties of the system using the equations of motion for a block and a spring.
First, we can find the maximum displacement of the block from equilibrium by using the equation for potential energy:
PE = (1/2)kx^2
where PE is the potential energy of the spring, k is the force constant, and x is the displacement. Plugging in the given values, we get:
PE = (1/2)(313 N/m)(0.230 m)^2 = 8.457 Nm
Next, we can use the conservation of mechanical energy to find the maximum speed of the block at the equilibrium position. The initial kinetic energy of the block is equal to the final potential energy of the spring:
(1/2)mv^2 = PE
Substituting the known values, we have:
(1/2)(2.10 kg)v^2 = 8.457 Nm
Simplifying the equation and solving for v, we find:
v = √(2PE/m) = √(2(8.457 Nm)/(2.10 kg)) ≈ 2.443 m/s
Therefore, the maximum speed of the block at the equilibrium position is approximately 2.443 m/s.