Question

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the spring constant?

Answer 1

`\textbf{2500 }\frac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}`

Step-by-step explanation:

Natural length of a spring is
`0.5\text{ }m`. The spring is streched by
`0.02\text{ }m`. The resultant energy of the spring is
`0.5\text{ }J`.

The potential energy of an ideal spring with spring constant
`k` and elongation
`x` is given by
`(1)/(2)kx^(2)`.

So, in the current problem, the natural length of the spring is not required to find the spring constant
`k`.

`\text{Potential Energy in the spring = }(1)/(2)kx^(2)\\0.5\text{ }J\text{ }=\text{ }(1)/(2)k(0.02\text{ }m)^(2)\\k*0.0004\text{ }m^(2)\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg(m^(2))/(s^(2))\\k\text{ }=\text{ }\frac{1\text{ }kg(m^(2))/(s^(2))}{0.0004\text{ }m^(2)}\text{ }=\text{ }2500\text{ }(kg)/(s^(2))`

∴ The spring constant of the spring =
`2500\text{ }(kg)/(s^(2))`