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3 years ago

A spring with a spring constant of 2.5 N/m is stretched by 2

Physics

1 answer:

mr_godi [17]3 years ago

3 0

Answer:

Eelas = 5 [J]

Explanation:

Elastic energy is associated with the ability that has a spring or any other material with elastic properties.

We can calculate the elastic energy by means of the following equation:

$E_{elas}=\frac{1}{2} *k*x^{2}$

where:

k = constant spring = 2.5 [N/m]

x = distance stretched = 2 [m]

$E_{elas}=\frac{1}{2} *2.5*(2)^{2}\\E_{elas}=5[J]$

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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 sp

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$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

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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 $\dfrac{1}{2}kx^{2}$ .

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$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

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