Calculating Proton Speed: 120V Potential Difference

[Gear300]

The question is to calculate the speed of a proton that is accelerated from rest through a potential difference of 120V.

Qp(charge of proton) ~ 1.60E-19C...with this, DeltaU = Qp*DeltaV...in which DeltaU is positive. It makes sense since the proton is heading towards a higher electric potential.

DeltaU = -DeltaK and since the proton accelerates from rest, DeltaU = -(1/2)*m*Vf^2 (Vf as final velocity). The problem I get here is that since DeltaU is positive, if I solve for Vf, I'll be square-rooting a negative number. I get the right answer if I ignore the negative sign, but otherwise, I get an imaginary number...what am I doing wrong?

Is it because of the DeltaU = -DeltaK...does that only hold valid for conservative forces? In this case, the rise in potential energy implies that the work is done by an external force.

 

[member 553083]

Is there a better way to calculate the speed of the proton in this case?The equation DeltaU = -DeltaK is only valid for conservative forces. In this case, the work done by the external force is equal to the change in kinetic energy of the proton. The equation to use in this case is W = DeltaK, where W is the work done by the external force. So, the equation to use to calculate the speed of the proton is W = (1/2)*m*Vf^2. Substituting W with Qp*DeltaV gives us Vf = sqrt[2*Qp*DeltaV/m]. Thus, the speed of the proton is: Vf = sqrt[2*(1.60E-19C)*(120V)/m], where m is the mass of the proton.