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

The distance is
`5.4*10^(-2)\ m`.

Step-by-step 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=(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=(1)/(2)kx^2`

`x^2=(2mc^2)/(k)`

Where, m = mass

c = speed of light

k = spring constant

Put the value into the formula

`x^2=(2*0.011*10^(-3)*(3*10^(2))^2)/(670)`

`x=√(0.002955)`

`x=0.054`

`x=5.4*10^(-2)\ m`

Hence, The distance is
`5.4*10^(-2)\ m`.