Question

A person brings a mass of $1kg$ from infinity to a point A. Initially, the mass at rest but it moves at a speed of $2m{s^{ - 1}}$ as it reaches A. The work done by the person on the mass is $- 3J$ .The gravitational potential at A is
(A) $- 3J/kg$
(B) $- 2J/kg$
(C) $- 5J/kg$
(D) None of the above.

Answer 1

Hint : The gravitational potential is potential energy per unit mass of a body. The work done will be equivalent to change in energy of a body as it moves from one point to another. So on finding the change in kinetic energy of the body and subtracting it from the work done, we will get the potential energy.

Formula Used: The formulae used in the solution are given here.
$\Rightarrow W = \Delta K.E. + \Delta P.E.$ where $W$ is the work done, $\Delta K.E.$ is the change in kinetic energy, and $\Delta P.E.$ is the change in potential energy.
$\Rightarrow K.E. = \dfrac{1}{2}m{v^2}$ where $m$ is the mass and $v$ is the velocity of the body.
$\Rightarrow P.E. = mgh$ where $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the vertical height from a chosen reference frame.
Gravitational potential $U = \dfrac{{\Delta P.E.}}{m}$

Complete step by step answer
Energy can be defined as the capacity for doing work. Work can be defined as transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. Work is the application of a force over a distance.
All forms of energy are either kinetic or potential. The energy in motion is known as kinetic energy $\left( {K.E.} \right)$ whereas potential energy $\left( {P.E.} \right)$ is the energy stored in an object and is measured by the amount of work done.
Thus, $W = \Delta K.E. + \Delta P.E.$ where $W$ is the work done, $\Delta K.E.$ is the change in kinetic energy, and $\Delta P.E.$ is the change in potential energy.
$\Rightarrow K.E. = \dfrac{1}{2}m{v^2}$ where $m$ is the mass and $v$ is the velocity of the body.
$\Rightarrow P.E. = mgh$ where $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the vertical height from a chosen reference frame.
Given that, mass $m = 1kg$ , work done $W = - 3J$ and $v = 2{m \mathord{\left/ {\vphantom {m s}} \right.} s}$ .
$\Rightarrow K.E. = \dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 1 \times {2^2}$
$\Rightarrow K.E. = 2J$
Substituting the values in the equation, $W = \Delta K.E. + \Delta P.E.$ , we get
$\Rightarrow - 3 = 2 + \Delta P.E.$
$\Rightarrow \Delta P.E. = - 5J$
The gravitational potential at a location is equal to the work done (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location.
Gravitational potential $U = \dfrac{{\Delta P.E.}}{m}$
Assigning the values obtained in the equation above we get,
$\Rightarrow U = - \dfrac{5}{1}J/kg = - 5J/kg.$
$\therefore$ The gravitational potential at A is $- 5J/kg.$
Therefore, the correct answer is Option C.

Note
Gravitational potential is also known as the Newtonian potential. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.