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= (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- (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- ((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.