Question

A person brings a mass of $1\text{kg}$ from infinity to a point $\text{A}$. Initially the mass was at rest but it moves at a speed of $2\text{m/s}$ as it reached $\text{A}$. The work done by the person on the mass is $-3\text{J}$. The potential at $\text{A}$ is

• A. $-3\text{J/kg}$
• B. $-12\text{J/kg}$
• C. $-5\text{J/kg}$
• D. None of these

Answer 1

The correct option is C $-5\text{J/kg}$

By work energy theorem, external workdone on the system is equal to the sum of change in potential and kinetic energy of system,
$$W_{ext}=\Delta U+\Delta K....\left(1\right)$$Given,
Mass, $m=1\text{kg}$

External work done by the person, $W_{ext}=-3\text{J}$

Initial velocity of mass , $v_{\text{initial}}=0$

Velocity of the mass at point $\text{A}$, $v_A=2\text{m/s}$

So, the change in kinetic energy,

$\Delta K=\frac{1}{2}m{v}_A^2-\frac{1}{2}m{v}_{\text{initial}}^2=\frac{1}{2}\times 1\times 2^2-\frac{1}{2}\times 1\times 0^2$.

$\Rightarrow \Delta K=2\text{J}$

Potential energy at infinity, $U_{\infty }=0$

Let, the potential energy at point $\text{A}$ be $U_A$.

Substituting the values obtained in equation $\left(1\right)$, we get

$-3=\Delta U+2$

$\Rightarrow \Delta U=-5\text{J}$

$\Rightarrow U_A-U_{\infty }=-5\text{J}$

$\Rightarrow U_A-0=-5\text{J}$

$\Rightarrow U_A=-5\text{J}$

The potential point $\text{A}$ is given by
$$V_A=\frac{U_A}{m}$$ $\Rightarrow V_A=\frac{-5}{1}$ $\Rightarrow \Delta V=-5\text{J/kg}$

Hence, option (c) is correct answer.

Key Concept: The change in potential is defined as the change in potential energy per unit mass. Why this question: To familiarize students with the relationship between potential energy and potential.