Question

What is the value of the spring constant of a spring with a potential energy of 8.67 J when it is
stretched 300 mm

Answer 1

Answer:

potential energy is told in a spring is given by U equals half gay X Square. Where is the spring grandstand on X? He's compression or extension in this particle. That problem the spring constant, is given by 27 Newton part meter, and the displacement is given by 16 centimeter, which is 0.16 major squared. Of that that comes out to be 13.5 times little 0.16 square. That comes out to be zero point 3456 Joon's this, but energies is stored as a potential energy when the spring is stretching.

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Answer 2

Answer:

When stretched by $300$mm, a spring with a potential energy of $8.67$j equaled under $92.66$, which is less than the $92.66$ value.

Explanation:

A spring's potential energy is equal to $8.67$joules when compressed to $300$ millimeters and then let to be $X$, which is equivalent to $300$ millimeters and further equals $0.3$ meters.

The spring constant's value must now be determined. The potential energy of a spring can be calculated using the well-known equation half times  $K(X)$square.  $8.67$j is half of the $K$ times  $0.3$squared in terms of all the values

Aware of  express $K$ as $8.67$ times $2$ divided by $0.3$ square, and then solve this further to find the solutions to $K$. Consequently, a spring constant's value would be less than $92.66$.

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