Question

A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.70 cm.
(a) How much energy has this system lost to damping during these eight cycles?
(b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

Answer 1

Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

ΔE (energy change) = Ef (final energy) - Ei (initial energy)

$\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J$

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.