Final answer:
The speed of the object when it is 8.00 cm from equilibrium is 0.50 m/s.
Step-by-step explanation:
To find the speed of the object when it is 8.00 cm from equilibrium, we can use the principles of simple harmonic motion. The potential energy stored in a spring is given by the equation U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium. In this case, the spring constant is 200 N/m and the displacement is 10.0 cm. Plugging in these values, we find that the potential energy stored in the spring when it is released is U = (1/2)(200)(0.10)^2 = 1.00 J.
The potential energy at any point in simple harmonic motion is equal to the kinetic energy. Therefore, when the object is 8.00 cm from equilibrium, it will have a kinetic energy of 1.00 J. The kinetic energy can be calculated using the equation KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity. Rearranging the equation gives v = sqrt(2KE/m). Plugging in the values, we find that v = sqrt(2(1.00)/(4.00)) = 0.50 m/s.