5.9 Relativistic energy  (Page 5/16)

Stored energy and potential energy

What happens to energy stored in an object at rest, such as the energy put into a battery by charging it, or the energy stored in a toy gun’s compressed spring? The energy input becomes part of the total energy of the object and thus increases its rest mass. All stored and potential energy becomes mass in a system. In seeming contradiction, the principle of conservation of mass (meaning total mass is constant) was one of the great laws verified by nineteenth-century science. Why was it not noticed to be incorrect? The following example helps answer this question.

Calculating rest mass

Strategy

Solution for (a)

1. Identify the knowns: $I·t=600\text{A}·\text{h;}V=12.0\text{V;}c=3.00\times {10}^8\text{m/s.}$
2. Identify the unknown: $\text{Δ}m.$
3. Express the answer as an equation:
   
   $\begin{array}{ccc}\hfill E_{\text{batt}}& =\hfill & (\Delta m)c^2\hfill \\ \hfill \Delta m& =\hfill & \frac{E_{\text{batt}}}{c^2}\hfill \\ & =\hfill & \frac{qV}{c^2}\hfill \\ & =\hfill & \frac{(It)V}{c^2}.\hfill \end{array}$
4. Do the calculation:
   
   $\text{Δ}m=\frac{(600\text{A}·\text{h})\left(12.0\text{V}\right)}{{(3.00\times {10}^8)}^2}.$
   
   Write amperes A as coulombs per second (C/s), and convert hours into seconds:
   
   $\begin{array}{cc}\text{Δ}m\hfill & =\frac{\left(600\text{C/s}·\text{h}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)(12.0\text{J/C)}}{{(3.00\times {10}^8\text{m/s)}}^2}\hfill \\ & =2.88\times {10}^{\mathrm{-10}}\text{kg.}\hfill \end{array}$
   
   where we have used the conversion $1\text{kg}·{\text{m}}^2{\text{/s}}^2=1\text{J}\text{.}$

Solution for (b)

1. Identify the knowns: $\text{Δ}m=2.88\times {10}^{\mathrm{-10}}\text{kg;}m=20.0\text{kg.}$
2. Identify the unknown: % change.
3. Express the answer as an equation: $\%\text{increase =}\frac{\text{Δ}m}{m}\times 100\%.$
4. Do the calculation:
   
   $\begin{array}{cc}\hfill \text{\% increase}& =\frac{\text{Δ}m}{m}\times 100\%\hfill \\ & =\frac{2.88\times {10}^{\mathrm{-10}}\text{kg}}{20.0\text{kg}}\times 100\%\hfill \\ & =1.44\times {10}^{\mathrm{-9}}\%.\hfill \end{array}$

Significance