Question

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 0.12 J. What is the greatest extension of the spring from its equilibrium length? What is the maximum speed of the block?

Answer 1

Answer:

The greatest extension of the spring is $\bf{0.055~m}$ and the maximum speed of the block is $\bf{0.695~m/s}$.

Explanation:

Given:

The mass of the block is, $m = 0.50~kg$

The spring constant of the spring is, $k = 80~N/m$

The mechanical energy of the block is, $E = 0.12~J$

When a particle is oscillating in a simple harmonic way, its total energy is given by

$E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

where $\omega$ is the angular velocity of the mass and $a$ is the amplitude of its motion.

The relation between angular velocity and spring constant is given by

$\omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting equation (2) in equation (1), we have

$~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Substituting $0.12~J$ for $E$ and $80~N/m$ for $k$ in equation (3), we can write

$a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m$

The relation between the maximum velocity and the amplitude is given by

$v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)$

Substituting $80~N/m$ for $k$ , $0.50~kg$ for $m$ and $0.055~m$ for $a$ in equation (4), we have

$v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s$