Energy Problem?

Question

1.1kg mass hangs from a spring with a force constant 400N/m. the mass is pulled down 13 cm from the equilibrium position and released. at the end of 5 complete cycles of vibration the mass reaches only 10cm from the equilibrium position.
a) how much mechanical energy is lost during these 5 cycles
b) what % of the mechanical energy is lost during the 5 cycles

Answer 1

For a spring system,

E = ½k A²


Ei = 0.5 (400) (.13)² = 3.38J

Ef = 0.5 (400) (.10)² = 2.00 J

a) Ef - Ei = 3.38 - 2 = 1.38 J
b) Ef / Ei = 59%

Answer 2

The first two answers neglect to take into account the potential energy of the mass (hanging from the spring), so you have to add to the 2J the change in potential energy caused by the fact that the mass is 3 cm higher than when released, which adds another 0.32J (m*g*h), assuming you take the initial position of the mass to be the reference height.

Answer 3

The energy stored in a spring is given by 0.5*k*s^2, where s = displacement and k = spring constant.

At the first pull, the energy is 0.5*k*(13cm)^2 = 0.5*k*(0.13m)^2

a the end of 5 cycles the energy is 0.5*k*(10cm)^2 = 0.5*k*(0.10m)^2

The difference is the energy lost during the 5 cycles

The % lost is the (difference divided by initial energy) x 100