Answer to Question #345972 in Mechanics | Relativity for larah

Question

How much work must be done to a proton to accelerate it from rest to 0.998c.

Answer 1

$$W=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2=mc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$$

$$W=1.67*10^{-27}*(3*10^8)^2(\frac{1}{\sqrt{1-0.998^2}}-1)\\ =2.2*10^{-9}\:\rm J$$