Question

Work done on a charge of mass $2kg$ , due to external force against electrostatic force is $- 10J$ if charge is displaced from A to B. Velocity of charge at point A is $4m/s$ and at B is $2m/s$ , then find the difference in potential energy $({u_B} - {u_A})$
(A) $+ 10J$
(B) $- 10J$
(C) $- 2J$
(D) $+ 2J$

Answer 1

Hint
To solve this question, we need to use the work energy theorem. We can find the kinetic energies of the particle at the two points from the respective velocities given. Then we need to put these values in the expression of the work energy theorem to get the answer.
The formula used in this solution is
$\Rightarrow W = \Delta K.E. + \Delta P.E.$ , here $W =$ work done by external force, $\Delta K.E.$ is the change in kinetic energy, and $\Delta P.E.$ is the change in potential energy.

Complete step by step answer
We know that the work energy theorem is given by
$\Rightarrow W = \Delta K.E. + \Delta P.E.$
Let ${K_A}$ and ${K_B}$ be the kinetic energies of the particle at A and B respectively.
$\therefore W = ({K_B} - {K_A}) + ({U_B} - {U_A})$
We know that $K = \dfrac{1}{2}m{v^2}$
So, the above expression can be written as
$\Rightarrow W = \dfrac{1}{2}m{v_B}^2 - \dfrac{1}{2}m{v_A}^2 + ({U_B} - {U_A})$
$\Rightarrow W = \dfrac{1}{2}m({v_B}^2 - {v_A}^2) + ({U_B} - {U_A})$
According to the question, $m = 2kg$ , ${v_A} = 4m/s$ , ${v_B} = 2m/s$ , and $W = - 10J$
Substituting these values in the above equation, we get
$\Rightarrow - 10 = \dfrac{1}{2}(2)({2^2} - {4^2}) + ({U_B} - {U_A})$
$\Rightarrow - 10 = ({2^2} - {4^2}) + ({U_B} - {U_A})$
On rearranging the terms, we get
$\Rightarrow ({U_B} - {U_A}) = {4^2} - {2^2} - 10$
$\Rightarrow ({U_B} - {U_A}) = 16 - 4 - 10$
Finally, we have the change in potential energy between A and B as
$\Rightarrow ({U_B} - {U_A}) = 2J$
So, the difference in the potential energies between the points A and B is $2J$ .
Hence, the correct answer is option (D), $+ 2J$.

Note
The work-energy theorem is more popularly known by the equation $W = \Delta K.E.$ but we have used here $W = \Delta K.E. + \Delta P.E.$ There should be no confusion between these two equations. In the first equation, $W$ refers to the work done by all the forces applied on the body. In the second equation, $W$ denotes the work done by the forces excluding the field forces applied on the body, such as the gravitational, electrostatic, spring forces etc. These are included in the $\Delta P.E.$ term. So, both of the equations are the same.