Answer:
1.32% error
Explanation:
You can derive the relativistic kinetic energy by integrating the velocity function over a differential change in momentum. Then use integration by parts to get the answer of:
[tex]KE=mc^2[gamma-1][/tex]
Gamma is the lorentz factor equal to:
[tex]gamma=\frac{1}{\sqrt{1-(v/c)^2}}[/tex]
Therefore the relativistic kinetic energy would be:
[tex]KE=mc^2[\frac{1}{\sqrt{1-(v/c)^2}}-1][/tex]
Plug everything in to get:
[tex]KE=(17000)(3x10^8)^2[\frac{1}{\sqrt{1-(0.14c/c)^2}}-1][/tex]
[tex]KE=(17000)(9x10^{16})[\frac{1}{\sqrt{1-0.0196}}-1][/tex]
[tex]KE=1.53x10^{21}[\frac{1}{\sqrt{1-0.0196}}-1][/tex]
[tex]KE=1.52x10^{19}J[/tex]
The classical non-relativistic formula for kinetic energy is:
[tex]KE=\frac{1}{2} mv^2[/tex]
Therefore the classical kinetic energy is:
[tex]KE=1.50x10^{19}J[/tex]
Use the error percentage formula, found online if you dont recall:
[tex]error=\frac{|KE_c-KE_r|}{KE_r} \frac{100}{1} =1.32[/tex]
