Consider a vertical spring with spring constant 28.43 N/m hanging from the ceiling. A small object with a mass of 1.281 kg is added to the spring and the spring stretches to its equilibrium position. The object is then pulled down a distance of 11.95 cm and released. What is the speed of the object a distance 8.927 cm from the equilibrium point?
We can start by finding the potential energy (PE) converted to kinetic energy (KE) as the object moves from the stretched position to the given distance from the equilibrium point.
At the stretched position:
PE_initial = (1/2) * k * (x_initial)^2
= (1/2) * 28.43 * (0.1195)^2
= 0.2028683525 J
At the given distance from the equilibrium point:
PE_final = (1/2) * k * (x_final)^2
= (1/2) * 28.43 * (0.08927)^2
= 0.11379616274 J
The difference in potential energy is converted to kinetic energy:
Delta_PE = PE_initial - PE_final
= 0.2028683525 - 0.11379616274
= 0.08907218976 J
Now we can find the speed of the object using the kinetic energy formula:
KE = (1/2) * m * v^2
0.08907218976 J = (1/2) * 1.281 kg * v^2
Solve for v:
v^2 = (2 * 0.08907218976 J) / 1.281 kg
v = sqrt((2 * 0.08907218976 J) / 1.281 kg)
v = 0.3944 m/s
The speed of the object at a distance of 8.927 cm from the equilibrium point is approximately 0.3944 m/s.