A spring is stretched from its equilibrium position 10 cm when a 1 kg mass is attached to it.Removing the 1 kg mass and replacing it with a 0.50 kg mass then releasing the spring,how high will the 0.50 kg mass rise above the point where it is released?
A) 0.44 m
B) 0.24 m
C) 0.80 m
D) 1.0 m

Question

Quizplus · Accepted Answer

The spring constant k can be found using Hooke's Law (F = kx) with the 1 kg mass: k = (1 kg * 9.8 m/s²) / 0.1 m = 98 N/m. When the 0.50 kg mass is attached, the potential energy stored in the spring is converted to gravitational potential energy at the highest point. Using energy conservation: (1/2)kx² = mgh, where x = 0.1 m, m = 0.50 kg, and g = 9.8 m/s². Solving for h gives h = 0.44 m.

A Spring Is Stretched from Its Equilibrium Position 10 Cm