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Suppose that the speed of light in a vacuum were one million times smaller than its actual value: c = 3.00 × 10^2 m/s. The spring constant of a spring is 670 N/m. Determine how far you would have to compress the spring from its equilibrium length in order to increase its mass by 0.011 g.

Respuesta :

Answer:

The distance is [tex]5.4\times10^{-2}\ m[/tex].

Explanation:

Given that,

Spring constant = 670 N/m

Mass = 0.011 g

We know that,

The potential energy stored in a compressed spring is given by

[tex]E=\dfrac{1}{2}kx^2[/tex]....(I)

We know that,

The equation of energy is

[tex]E = mc^2[/tex]....(II)

We need to calculate the distance

Using equation (I) and (II)

[tex]mc^2=\dfrac{1}{2}kx^2[/tex]

[tex]x^2=\dfrac{2mc^2}{k}[/tex]

Where, m = mass

c = speed of light

k = spring constant

Put the value into the formula

[tex]x^2=\dfrac{2\times0.011\times10^{-3}\times(3\times10^{2})^2}{670}[/tex]

[tex]x=\sqrt{0.002955}[/tex]

[tex]x=0.054[/tex]

[tex]x=5.4\times10^{-2}\ m[/tex]

Hence, The distance is [tex]5.4\times10^{-2}\ m[/tex].

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