Respuesta :
Answer:
The greatest extension of the spring is [tex]\bf{0.055~m}[/tex] and the maximum speed of the block is [tex]\bf{0.695~m/s}[/tex].
Explanation:
Given:
The mass of the block is, [tex]m = 0.50~kg[/tex]
The spring constant of the spring is, [tex]k = 80~N/m[/tex]
The mechanical energy of the block is, [tex]E = 0.12~J[/tex]
When a particle is oscillating in a simple harmonic way, its total energy is given by
[tex]E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]\omega[/tex] is the angular velocity of the mass and [tex]a[/tex] is the amplitude of its motion.
The relation between angular velocity and spring constant is given by
[tex]\omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Substituting equation (2) in equation (1), we have
[tex]~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Substituting [tex]0.12~J[/tex] for [tex]E[/tex] and [tex]80~N/m[/tex] for [tex]k[/tex] in equation (3), we can write
[tex]a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m[/tex]
The relation between the maximum velocity and the amplitude is given by
[tex]v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)[/tex]
Substituting [tex]80~N/m[/tex] for [tex]k[/tex] , [tex]0.50~kg[/tex] for [tex]m[/tex] and [tex]0.055~m[/tex] for [tex]a[/tex] in equation (4), we have
[tex]v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s[/tex]
