whats the relation between Kinetic Energy and Potential difference?

Question

i just need to know the relation in order to solve the following question;
A proton is accelerated from rest through a potential difference of 25700 V

a) whats the kinetic energy of this proton in joules after this acceleration?

b) whats the speed of the proton after this acceleration?

ok now, i got how to get the E.P.E. (Electrical Potential Energy ) from Potential difference and the charge, but the PROBLEM IS , HOW TO CONVERT THE E.P.E TO K.E. ??? PLZ EXPLAIN

Answer 1

________________________________________
The ratio of workdone(w) to charge(q) is electric potential difference(V)

w =qV

Kinetic energy =KE=work(w)

KE=qV

charge of proton = q =1.6*10^-19 C

potential difference =V = 25700 V

The kinetic energy of the proton after this acceleration= [1.6*10^-19 ]*[25700]=4.112*10^-15 J

a) The kinetic energy of the proton after this acceleration= 4.112*10^-15 J

__________________________________

The speed of the proton after acceleration = v = sq rt [2KE/m]

v=sq rt 2*4.112*10^-15/1.67*10^-27

v = sq rt 8.224*10^-15/1.67*10^-27

v = 2.22*10^6 m/s

(b)The speed of the proton after acceleration is2.22*10^6 m/s
____________________________________
_________________________________________

Answer 2

Potential And Kinetic Energy Equation

Answer 3

Energy and work both have the same units, Nm or Joules. Work is the transformation of energy from one form to another. When you push an object along the ground, the work done is equal to the force (weight x coefficient of friction) times the distance. The energy used in pushing the object is converted into heat. If you lift a mass m a distance h, the work done is mgh (mg is the force, h the distance). The lifting energy is converted into potential energy. When the object falls, its velocity after traveling distance h is sqrt(2gh), and its kinetic energy is 1/2 mv^2 = 1/2 m * 2gh = mgh, showing that the work, the kinetic energy, and the potential energy are indeed of identical value.

Answer 4

A potential difference is defined as the potential energy per unit charge associated with an electric field. I.e., V = E/q. Thus, E = q*V is the change in potential energy caused by a charge q moving through a potential difference V. But due to conservation of energy, E would also be the change in kinetic energy (KE). Since we're starting from rest, we can say that KE = q*V. You'll need to know the charge of a proton to find the answer to a), and the mass of a proton to find the answer to b).

Answer 5

If energy is conserved (and it is in this case), then the sum of the potential energy and the kinetic energy is a constant.

Now the proton starts from rest so it's kineitc energy is zero, but it has a potential energy, PE0 equal to:

PE0 = qV where q =1.6x10^-19 Coul and V = 27,500 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE1 = 0, KE1 = 1/2mv^2

Now since PE0 +KE0 = Total energy =PE1 + KE1

qV + 0 = 0 + 1/2mv^2

Or: KE = qV = 4 x 10^-15 J (Part A)

The speed of the proton is just

v = sqrt(2qV/m) = sqrt(2*PE0/m) = sqrt(2*4x10^-15J/(1.6x10^-27 kg))

v = 2.2 x 10^6 m/s