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

`(E)/(E_c) =3.125`

Step-by-step explanation:

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

`E_c=(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* 10^(8) m/s`

Hence:

`E_c=(1)/(2) (74.7)*(0.8*3* 10^(8) )^2=2.15136* 10^(18) J`

`E=(74.7)*(3* 10^(8) )^2 =6.723* 10^(18) J`

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

`(E)/(E_c) =(6.723* 10^(18))/(2.15136* 10^(18)) =3.125`