Final answer:
Using conservation of energy, the speed of the 1.00-kg mass 1.00 cm above the release point on a spring with a constant of 1.00 N/cm is approximately 0.316 m/s.
Step-by-step explanation:
The question asks about the speed of a mass when it is 1.00 cm above the point from which it was released, after being pulled downward 2.00 cm on a spring with a spring constant of 1.00 N/cm. We can solve this problem using the principle of conservation of energy. In this case, the total mechanical energy (potential energy in the spring plus kinetic energy) is conserved because no non-conservative forces (like friction) are doing work.
At the starting point, 2.00 cm below the equilibrium, all of the energy is in the form of potential energy stored in the spring:
PEs = 1/2 k x2, where k is the spring constant and x is the displacement from equilibrium.
PEs = 1/2 (1.00 N/cm)(2.00 cm)2 = 0.0200 J
When the mass passes the point 1.00 cm above the release point, some potential energy has been converted into kinetic energy. We can find the velocity by setting the potential energy at this point equal to the initial potential energy minus the kinetic energy.
PEs,final + KE = PEs,initial
1/2 k x2 + 1/2 m v2 = 1/2 k (2.00 cm)2
Substituting values we get:
1/2 (1.00 N/cm)(1.00 cm)2 + 1/2 (1.00 kg)v2 = 0.0200 J
Solving for v gives the speed of the mass at 1.00 cm above the release point. Carrying out the algebra yields:
v = sqrt(2(0.0200 J - 1/2 (1.00 N/cm)(1.00 cm)2)/(1.00 kg))
v ≈ 0.316 m/s
The speed of the mass 1.00 cm above the release point is approximately 0.316 m/s.