How Much Work to Accelerate a Proton Between Specified Speeds?

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SUMMARY

The discussion focuses on calculating the work required to accelerate a proton from specified speeds, specifically from 0.15c to 0.16c, 0.81c to 0.82c, and 0.93c to 0.94c. The mass of the proton is given as 1.67x10^-27 kg, and the energy is expressed in MeV. The initial approach using the equation e=mc^2 is incorrect for these calculations; instead, the kinetic energy formula E = 1/2 mv^2 should be used, with a relativistic correction for speeds approaching the speed of light.

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patapat
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Homework Statement


How much work must be done on a proton to increase its speed by each of the values below?

(a) 0.15c to 0.16c
(b) 0.81c to 0.82c
(c) 0.93c to 0.94c

The answer is given in MeV which is a million electron volts.
mass of proton=1.67x10^-27kg

Homework Equations


e=mc^2
1 J=1.6x10^-19eV=1.6x10^-13MeV

The Attempt at a Solution


I'm not sure if this is even the correct equation to use, but I don't see any other appropriate equations in this book.

Energy required at .15c
e=(1.67x10^-27kg)(.15*3x10^8)^2
e=3.38x10^-12J
Convert to MeV
e=21.14MeV

Energy required at .16c
e=(1.67x10^-27kg)(.16*3x10^8)^2
e=3.85x10^-12J
convert to MeV
e=24.05MeV

Energy required to go from .15c-16c
24.05MeV-21.14MeV=2.91MeV
Which I found to be wrong, obviously.

Thanks in advance to any and all help. Much appreciated.

-Pat
 
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You have misunderstood the equation, in e=mc^2 C is the speed of light and is always constant. The value s0.15C etc that you have been given are speeds as fractions of the speed of light.
Kinetic energy is E = 1/2 m v^2, there is a small correction for relativistic speeds which you will have to lookup for the higher spped cases.
 

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