Question

How much work is required to accelerate a proton from rest up to a speed of 0.987 c?

Answer 1

For the work-energy theorem, the work done to accelerate the proton is equal to the variation of kinetic energy of the proton:
$W= \Delta K = K_f - K_i =K_f$
where $K_i$ is the initial kinetic energy, which is zero because initially the proton is at rest.

We need to find the kinetic energy of the proton: we can't use the classical formula $K= \frac{1}{2} mv^2$ since the proton is at relativistic speed. Therefore, we must use the relativistic formula:
$K_f = \frac{mc^2}{\sqrt{1- \frac{v^2}{c^2} }}-mc^2$
where 
m is the proton mass
v is its final speed
c is the speed of light

Substituting $m=1.67 \cdot 10^{-27} kg$, v=0.987 c and $c=3 \cdot 10^8 m/s$, we find
$K_f = \frac{(1.67 \cdot 10^{-27} kg)(3 \cdot 10^8 m/s)^2}{\sqrt{1- \frac{(0.987 c)^2}{c^2} }} -(1.67 \cdot 10^{-27}kg)(3 \cdot 10^8 m/s)^2=$
$=7.82 \cdot 10^{-10} J$
and this is equal to the work needed to accelerate the proton up to a speed of 0.987c.