Solve Proton Bombardment Homework: Find Vf in m/s

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Homework Statement


A proton with mass 1.45*10^-27 kg is propelled at an initial speed of 3.99*10^5 m/s directly toward a uranium nucleus 5.16 m away. The proton is repelled by the uranium nucleus with a force of magnitude F = a/x^2 where x is the separation between the two objects and a = 2.12 *10^-26 Nm2. Assume that the uranium nucleus remains at rest. What is the speed of the proton when it is 8.79 *10^-10 m from the nucleus?

Give your answer in m/s in scientific notation to three significant digits (i.e. 1.15*10^2)

Note: The numbers given in this problem do not represent actual figures, instead this is to be thought of as a theoretical calculation.



Homework Equations


F = a/x^2
W = F*D
KE = .5*m*Vf^2-.5*m*Vi^2

The Attempt at a Solution



F = a/x^2
F = 2.12 *10^-26/(8.79 *10^-10)^2 = 2.7438*10^-8 N
W = F*D
W = 2.7438*10^-8*(5.16 - 8.79 *10^-10) = 1.4158*10^-7 J
KE = .5*m*Vf^2-.5*m*Vi^2
1.4158*10^-7 = .5(1.45*10^-27)Vf^2-(.5)(1.45*10^-27)(3.99*10^5)^2
Vf^2 = 1.95285*10^20
Vf = 1.397410*10^10 m/s

The correct answer is supposed to be 3.55*10^5 m/s, no matter how many times I try to work this problem out I cannot every arrive at the correct value for Vf.
 
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You have to integrate the force eqn wrt dist to find the work done. The force is a function of dist, not constant.

Otherwise, use the concept of PE and KE which you must have learned in electrostatics.