Energy of block on a spring
- 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