1. Learning Object
Energy of a Block on a Spring
Given that when a particular block of 25 grams oscillates with a maximum displacement
of 12 cm from the equilibrium point, determine both the total energy for the system and
the block’s velocity at 4 cm. above the equilibrium point.
Hint: Think about the block’s kinetic energy at 12 cm, and what the total energy must be
a combination of.

2. Solution:
Since total energy is just a sum of kinetic energy and potential energy, set up the
equation with mass as 25, velocity as 0, and 0.12 meters as the height. Then solve for
total energy.
E = KE + PE
E = 0.5mv^2 + mgh
E = 0.5(25)(0^2) + 25(9.8)(0.12)
E = 29.4 J
When the block is 4 cm above the equilibrium height, you substitute 0.04 in for h. The
entire equation must always equal 29.4 because of the conservation of energy, so solve
for v in the equation.
29.4 = 0.5(25)v^2 + 25(9.8)(.04)
29.4 = 12.5v^2 + 9.8
19.6 = 12.5v^2
v^2 = 1.568
v = 1.252 m/s