A $1.00 - \mathrm { kg }$ gold ingot is carried from the bottom of the Eiffel tower to the top, $300 \mathrm {~m}$ above ground level. By what amount does its mass change? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } \right)$
A) Its mass decreases by $3.27 \times 10 ^ { - 14 } \mathrm {~kg}$
B) Its mass decreases by $1.64 \times 10 ^ { - 14 } \mathrm {~kg}$
C) Its mass increases by $3.27 \times 10 ^ { - 14 } \mathrm {~kg}$
D) Its mass does not change.
E) Its mass increases by $1.64 \times 10 ^ { - 14 } \mathrm {~kg}$

Question

Quizplus · Accepted Answer

The potential energy gained by the gold ingot is converted into an equivalent mass increase according to $E=mc^2$. The increase in potential energy is $mgh = 1 \times 9.8 \times 300 = 2940 \, \text{J}$. The mass increase is $\Delta m = \frac{E}{c^2} = \frac{2940}{(3 \times 10^8)^2} \approx 3.27 \times 10^{-14} \, \text{kg}$.

A 1.00−kg1.00 - \mathrm { kg }1.00−kg Gold Ingot Is Carried from the Bottom of the Eiffel

A $1.00 - \mathrm { kg }$ Gold Ingot Is Carried from the Bottom of the Eiffel