Question

If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy?

Answer 1

Answer:

$\frac{E}{E_c} =3.125$

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

$E_c=\frac{1}{2} mv^2$

The equivalence between mass and energy established by the theory of relativity is given by:

$E=mc^2$

This formula states that the equivalent energy $E$ can be calculated as the mass $m$ multiplied by the speed of light $c$ squared.

Where $c$ is approximately $3\times 10^{8} m/s$

Hence:

$E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J$

$E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J$

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

$\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125$