A 1.33 kg object is attached to a horizontal spring of force constant 2.50 N/cm and is started oscillating by pulling it 6.40 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.

Relax

Respuesta :

Answer:

Explanation:

a ) Energy of spring = 1/2 k A² where A is amplitude of oscillation and k is force constant .

So initial energy = 1/2 x 2.5 x (6.4 x 10⁻²)²

= 51.2 x 10⁻⁴ J

So final  energy = 1/2 x 2.5 x (3.7 x 10⁻²)²

= 17.11 x 10⁻⁴ J

energy lost

= 34.1 J .

This energy is dissipated in the form of heat,  sound etc.

Answer:

a. [tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

b. The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.

Explanation:

Given:

  • mass of the attached object, [tex]m=1.33\ kg[/tex]
  • spring constant, [tex]k=2.5\ N.cm^{-1}[/tex]
  • maximum displacement, [tex]A=6.4\ cm[/tex]
  • maximum displacement after damping, [tex]x=3.7\ cm[/tex]

a)

the energy lost in damping:

[tex]\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2[/tex]

[tex]\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)[/tex]

[tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

b)

The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.