Physics, 27.06.2019 18:50 jljhenkel

16. a 6 kg block is falling toward a spring located 1.8 m below. if it has a speed of 4 m/s at first, what will be the maximum compression of the spring? assume the spring constant is 2000 n/m[6 marks][1 mark for comm]

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Answer from: evalentine843

Maximum compression of the spring: $\rm 0.423\; m$ .

Explanation:

Let be the maximum compression, in meters, of the spring.

What's the initial kinetic energy of this block?

The initial velocity

of the block is $\rm 4\;m\cdot s^{-1}$ .The mass of the block is $\rm 6\;kg$ .Initial $\displaystyle \mathrm{KE} = \frac{1}{2} \;m\cdot v^{2} = \rm \frac{1}{2} \times 6\times 4^{2} = 48\;J$ .

What's the final kinetic energy of this block?

The block would have come to a stop by the time the spring reaches its maximum compression. v = 0

. As a result, the final kinetic energy of the block shall also be zero.

Let the initial height of the block be zero.

The height of the block will be $\rm -1.8\;m$ by the time it reaches the top of the spring. However, the height of the block will become even more negative as the spring compresses. If the maximum compression of the spring is meters, the height of the block will be -(1.8 + x) meters by the time the spring reaches its maximum compression.

The change in the height of the block will thus be $-(1.8 + x)\;\mathrm{m}$ relative to the initial position of the block. The change in the block's potential energy due to gravity will be:

$\Delta \mathrm{GPE} = m\cdot g\cdot \Delta h = 6\times 9.8 \times (-(1.8 + x)) = -105.948- 58.86\;x$ .

The mechanical energy of an object is the sum of its kinetic energy and its potential energy. What will be the change in the mechanical energy of the block?

$\text{Change in Mechanical Energy} \\ = \text{Change in KE} + \text{Change in PE} \\ = (0 - 48) + ((-105.948- 58.86\;x) - 0)\\ = (-153.948 - 58.86\;x)\; \text{J}$ .

The block has lost $(153.948 + 58.86\;x)\; \text{J}$ of energy as it travels from its initial position to the place where the spring is most compressed. Energy conserves. Those energies are transferred to the spring. Assume that there's no energy loss in this process. All those $(153.948+ 58.86\;x)\; \text{J}$ would have become the elastic potential energy of the spring.

On the other hand, it is assumed that the spring is compressed by a distance of meters. By Hooke's Law, the elastic potential energy of the spring will be:

$\displaystyle \text{Elastic Potential Energy} = \frac{1}{2} \; k\cdot x^{2} = \frac{1}{2}\times 2000 \times x^{2} = 1000\; x^{2}\;\text{J}$ .

There are thus two expressions for the Elastic Potential Energy of the spring:

$(153.948+ 58.86\;x)\; \text{J}$ from the conservation of energy, and $1000\; x^{2}\;\text{J}$ by Hooke's Law.

The two expression shall be equivalent. Equate the two expressions and solve for the compression of the spring, .

$153.948+ 58.86\;x = 1000\; x^{2}$ .

$1000\; x^{2}- 58.86\;x - 153.948 = 0$ .

$\displaystyle x_{1} = \frac{-(-58.86) +\sqrt{(-58.86)^{2} - 4\times 1000 \times (-153.948)}}{2\times 1000} = \rm 0.423$ .

$\displaystyle x_{2} = \frac{-(-58.86) +\sqrt{(-58.86)^{2} - 4\times 1000 \times (-153.948)}}{2\times 1000} = -0.364$ .

Consider the expression for the height of the block when it is at the top of the spring and when the spring reaches maximum compression:

Height of the block at the top of the spring: $\rm -1.8\;m$ ;Height of the block when the spring reaches maximum compression: $-(1.8 + x) \; \text{m}$ . This height will be $\rm -2.2\; m$ if $x_{1}$ is correct, and $\rm -1.4\; m$ if $x_{2}$ is correct. The block compresses the spring and will be below its initial position at the top of the spring. Therefore, x 0

and only $x_{1}$ is correct.

The maximum compression of the spring will be 0.423 meters.

Answer from: Quest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

Answer from: Quest

the earths atmosphere is made up of nitrogen and oxygen. solids can turn into gasses through a process called sublimation. sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

hope this : d