Determine the fractional change in the mass of a spring upon being compressed by 1 cm . At its natural length, the spring has a mass of 200 g and a spring constant of 500 N m^ -1 .

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Correct answer: C. 1.4 × 10^ -18

Energy stored in the spring is given by

U = \dfrac{1}{2}kx^2

.

Substituting the given values:

U = \dfrac{1}{2} \times 500 \times (0.01)^2 = 0.025 \text{ J}

By mass-energy equivalence, the change in mass is

\Delta m = \dfrac{U}{c^2}

.

The fractional change in mass is

\dfrac{\Delta m}{m} = \dfrac{U}{mc^2}

.

Substituting the values:

\dfrac{\Delta m}{m} = \dfrac{0.025}{0.2 \times (3 \times 10^8)^2}

\dfrac{\Delta m}{m} = \dfrac{0.025}{0.2 \times 9 \times 10^{16}} = \dfrac{0.025}{1.8 \times 10^{16}}

\dfrac{\Delta m}{m} = \dfrac{2.5}{1.8} \times 10^{-18} \approx 1.388 \times 10^{-18} \approx 1.4 \times 10^{-18}

Determine the fractional change in the mass of a spring upon being...

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