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A proton with a speed of 4.0*10^4 m/s falls through a potential difference V and thereby increases its speed to 9.0*10^4 m/s. Through what potential

A proton with a speed of 4.0*10^4 m/s falls through a potential difference V and thereby increases its speed to 9.0*10^4 m/s. Through what potential difference did the proton fall? Answer is 34V but not sure how I got it. I used conservation of energy but keep getting 3.34E-37

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