Question

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.

Answer 1

Answer:

The distance is $5.4\times10^{-2}\ m$.

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

$E=\dfrac{1}{2}kx^2$....(I)

We know that,

The equation of energy is

$E = mc^2$....(II)

We need to calculate the distance

Using equation (I) and (II)

$mc^2=\dfrac{1}{2}kx^2$

$x^2=\dfrac{2mc^2}{k}$

Where, m = mass

c = speed of light

k = spring constant

Put the value into the formula

$x^2=\dfrac{2\times0.011\times10^{-3}\times(3\times10^{2})^2}{670}$

$x=\sqrt{0.002955}$

$x=0.054$

$x=5.4\times10^{-2}\ m$

Hence, The distance is $5.4\times10^{-2}\ m$.