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A person holds a 4.0 kg block at position A shown above on the left, in contact with an uncompressed vertical spring with a spring constant of 500 N/m. The person gently lowers the block from rest at position A to rest at position B. Which of the following describes the change in the energy of the block-spring-Earth system as a result of the block being lowered?

A Person Holds A 40 Kg Block At Position A Shown Above On The Left In Contact With An Uncompressed Vertical Spring With A Spring Constant Of 500 Nm The Person G class=

Sagot :

NusrathC

Answer:

Energy decrease by approximately 1.5J

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The change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.

Given information;

A person holds a 4.0 kg block at position A in contact with an uncompressed vertical spring with a spring constant, k, of 500 N/m.

The person gently lowers the block from rest at position A to rest at position B. The decrease in height, h, of the object or the compression in spring, x, will be 10 cm.

Due to the decrease in height of the block, the decrease in potential energy will be,

[tex]PE_b=-mgh\\=-4\times9.81\times0.1\\=-3.924\rm\;N[/tex]

Due to compression, the increase in potential energy of the spring will be,

[tex]PE_s=\dfrac{1}{2}kx^2\\=\dfrac{1}{2}500\times(0.1)^2\\=2.5\rm\;J[/tex]

So, the net change in the energy of the system will be,

[tex]E=-3.924+2.5\\=1.424\rm\;J[/tex]

Therefore, the change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.

For more details, refer to the link:

https://brainly.com/question/2730954

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