How Much Kinetic Energy to Accelerate a Proton to 0.9999c?

[StillLearningToronto]

Question:

Calculate the kinetic energy required to accelerate a single proton from rest position to 0.9999c. The mass of the proton is 1.67*10^-27 kg.

Equations Used:
Ek=(mc^2/√ 1-(v^2/c^2))-mc^2

Attempt at the problem:

Ek=(1.67*10^-27kg)(3*10^8m/s)^2/√ 1-(0.9999c^2/c^2))-1.67*10^-27kg)(3*10^8m/s)^2
Ek=(1.5*10^-10/0.01)-1.5*10^-10
Ek= 1.48*10^-8

(sorry if its jumbled, I am used to writing it out in my equation editor)

I'm not fairly confident on this answer because I have found different answers online saying it was correct.

Just looking for someone to verify this, or at least point me in the right direction,
Cheers xx

 

[JoePhysics]

I didn't check your arithmetic, but an answer like "Ek= 1.48*10^-8" can never be right. Are you sure you aren't forgetting something?

 

[StillLearningToronto]

I didn't check your arithmetic, but an answer like "Ek= 1.48*10^-8" can never be right. Are you sure you aren't forgetting something?

Im very positive I am missing something.

This is what i found online:

Ekrest = mc2
Ekrest = (1.67*10^-27)(c)2
Ekrest = 1.5*10^-10

Ek= (1.5*10^-10)/√(1.9999*10^-4)-(1.5*10^-10)
EK=1.047*10^-8J

But i have NO idea how they got (1.9999*10^-4)

 

[robphy]

But i have NO idea how they got (1.9999*10^-4)

Are you sure that everything is properly "squared" away?

 

[kuruman]

But i have NO idea how they got (1.9999*10^-4)

I too have no idea how they got this number or what it means, but their numerical answer 1.047×10^-8 J agrees with what I got. So let me guide you through the steps that I took.
1. Write an expression for the total energy of the proton using the Lorentz factor γ.
2. Subtract the rest mass energy to find the kinetic energy of the proton.
3. Put in the numbers.

 

[robphy]

http://m.wolframalpha.com/input/?i=(proton+mass)*(cosh(arctanh(.9999))-1)c^2+in+joules+&x=0&y=0

http://m.wolframalpha.com/input/?i=1/cosh(arctanh(.9999))^2&x=0&y=0

http://m.wolframalpha.com/input/?i=1-.9999^2&x=0&y=0