Physics, 26.06.2020 16:01, hernandez09297

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring

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answered: whitty

a) The spring constant of the spring is $28.381\,\frac{N}{m}$ , b) The angular frequency of the block is $7.038\,\frac{rad}{s}$ .

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

$m\cdot g = k\cdot \Delta x$

Where:

- Mass, measured in kilograms.

- Gravitational constant, measured in meters per square second.

- Spring constant, measured in newtons per meter.

$\Delta x$ - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ( $m = 0.573\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta x = 0.198\,m$ )

$k = \frac{m\cdot g}{\Delta x}$

$k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}$

$k = 28.381\,\frac{N}{m}$

The spring constant of the spring is $28.381\,\frac{N}{m}$ .

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

$\omega = \sqrt{\frac{k}{m} }$

Given that $k = 28.381\,\frac{N}{m}$ and $m = 0.573\,kg$ , the angular frequency, measured in radians per second, of the block is:

$\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }$

$\omega = 7.038\,\frac{rad}{s}$

The angular frequency of the block is $7.038\,\frac{rad}{s}$ .

answered: Guest

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answered: Guest

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this. hope that you like the video and that it explains more in detail to you.

that is the link.

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A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring....