Two carts connected by a 0.25 m spring hit a wall, compressing the spring to 0.1 m. The spring constant k is 100 N m . What is the elastic potential energy stored from the spring's compression?

Answers

Answer 1

Given that,

Initial position of the spring, x₁ = 0.25 m

Final position of the spring, x₂ = 0.1 m

The spring constant of the spring, k = 100 N/m

To find,

The elastic potential energy stored from the spring's compression.

Solution,

The work done in moving a spring from on position to another is stored in the form of elastic potential energy. It can be given as follows :

[tex]E=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 100\times (0.1^2-0.25^2)\\\\=2.62\ J[/tex]

So, the required energy is 2.62 J.

Answer 2

Answer:

1.125 J

Explanation:

The correct answer is 1.125 J


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For a given force as time ___
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Answers

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Explanation:

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