Question

An ideal spring hangs from the ceiling. A 1.55 kg1.55 kg mass is hung from the spring, stretching the spring a distance ????=0.0855 md=0.0855 m from its original length when it reaches equilibrium. The mass is then lifted up a distance ????=0.0235 mL=0.0235 m from the equilibrium position and released. What is the kinetic energy of the mass at the instant it passes back through the equilibrium position?

Answer 1

Answer: 0.048 J

Explanation:

The described situation is better understood with the attached figure.

Let's assume that when the mass is released after being lifted up, it starts performing simple harmonic motion with an amplitude L. Then, the maximum speed $V$ of this hanging mass is fulfilled at the equilibrium position and its given by the following equation:

$V=\sqrt{\frac{k}{m}}L$ (1)

Where:

$k$ is the spring constant which can be calculated by the Hooke's law: $k=\frac{F}{d}=\frac{mg}{d}$ being $g=9.8 m/s^{2}$ the acceleration due gravity and $d=0.0855 m$ the length the spring is streched.

$m=1.55 kg$ is the mass

$L=0.0235 m$ is the amplitude

So, $k=\frac{(1.55 kg)(9.8 m/s^{2})}{0.0855 m}=177.66 N/m$ (2)

Substituting (2) in (1):

$V=\sqrt{\frac{177.66 N/m}{1.55 kg}}0.0235 m$ (3)

$V=0.251 m/s$ (4)

On the other hand, the kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mV^{2}$ (5)

$K=\frac{1}{2}(1.55 kg)(0.251 m/s)^{2}$ (6)

Hence:

$K=0.048 J$