Q64PE - College Physics (Urone)

Question

(a) Calculate the energy released by the destruction of 1.00kg of mass. (b) How many kilograms could be lifted to a 10.0 km height by this amount of energy?

Step-by-Step Solution

Answer

1. The energy released is obtained as E = 9.00 ×10¹⁶ J .
2. The weight lifted is obtained as m = 9.18 ×10¹¹ kg .

Step 1: Given Data

The amount of mass destructed is: 1.00kg

The distance is 10.0km.

Step 2: Define Relativity

Regarding the nature and behavior of light, space, time, and gravity, relativity is the idea that different physical events rely on the relative motion of the observer and the observed objects.

Step 3: Evaluating the energy released

a. We know that the rest mass energy is given by:

\[E{\rm{ }}={\rm{ }}m{\rm{ }}{c^2}\]………………(I)

Here the value of \[m\] is the mass of the object.

The value \[c{\rm{ }}={\rm{  }}3{\rm{ }}x{\rm{ }}{10^8}{\rm{ }}m/s\] is the speed of light.

The mass of the material then is:

\[m=1.00{\rm{ }}kg\]

With the help of the first equation, the rest mass energy of the \[1.00{\rm{ }}kg\] mass is obtained as:

\begin{aligned}E{\rm{ }}={\rm{ }}m{\rm{ }}{c^2}\\=1.00{\rm{ }}kg \times {(3{\rm{ }}x{\rm{ }}{10^8}{\rm{}}m{\rm{}}{s^{-1}})^2}\\= 9.00 \times {10^{16}}{\rm{ }}kg{\rm{ }}{m^2}{\rm{ }}{s^{ - 2}}\\=9.00 \times {10^{16}}J \end{aligned}

Therefore, the energy released is:9.00 ×10¹⁶ J. \[9.00 \times {10^{16}}J\].

Step 4: Evaluating the weight lifting

b. As we know the potential energy is obtained by:

\[PE{\rm{ }}={\rm{ }}m{\rm{ }}g{\rm{ }}h\]………………(II)

Here the value of \[m\] is the mass of the material,

The value of \[h\] is the height.

The value of \[g{\rm{ }} = {\rm{ }}9.8{\rm{ }}m{s^{ - 2}}\] is said to be the gravitational acceleration.

It given that the height upto which material to be lifted is:

\begin{aligned}h = 10.0km\\=10.0 \times {10^3}m \end{aligned}

Here it is given that the energy calculated in the part (a) is stored in the material as the potential energy.

Then with the help of the second equation, we get:

\begin{aligned}m=\frac{{PE}}{{gh}}\\=\frac{{9.00 \times {{10}^{16}}{\rm{ }}J}}{{9.8{\rm{ }}m{s^{-2}}\times10.0\times{{10}^3}{\rm{}}m}}\\=9.18\times{10^{11}}{\rm{}}kg\end{aligned}

Therefore, the weight lifted is 9.18 ×10¹¹ kg \[9.18 \times {10^{11}}{\rm{ }}kg\].