Question

A body is dropped from a height $h$. When the loss in its potential energy is $U$ then its velocity is $v$. The mass of the body is:
A) $\dfrac{{{U^2}}}{{2v}}$
B) $\dfrac{{2v}}{U}$
C) $\dfrac{{2v}}{{{U^2}}}$
D) $\dfrac{{2U}}{{{v^2}}}$

Answer 1

Hint The law of conservation of energy relates the body’s kinetic and potential energy. When the body is at a height $h$, it will only have potential energy and when it's falling and has lost potential energy $U$ it will have some kinetic energy. We will use this relation to equate the loss in potential energy of the body with the gain in kinetic energy of the body as it accelerates downwards.
In this solution, we will use the following formula
Kinetic energy of a body: $K = \dfrac{1}{2}m{v^2}$ where $m$ is its mass and $v$ is its velocity.

Complete step by step answer
The law of conservation of energy tells us that energy cannot be created or destroyed but it can only be converted from one form to the other.
We’ve been given that a body is dropped from a height $h$. When the loss in its potential energy is $U$, the amount of energy will have converted into kinetic energy. We know that the kinetic energy of a body travelling with velocity $v$ and mass $m$ can be calculated as:
$\Rightarrow K = \dfrac{1}{2}m{v^2}$
So, the loss in the potential energy $U$ is converted into kinetic energy so we can write
$\Rightarrow U = \dfrac{1}{2}m{v^2}$
Dividing both sides by ${v^2}/2$, we get
$m = \dfrac{{2U}}{{{v^2}}}$ which corresponds to option (D).

Note
Here we have assumed that there is no air resistance however in an actual scenario, air resistance will decrease the kinetic energy of the body due to frictional losses. We can also select the correct option since only option (D) has the dimensions of mass while the rest of the options have different dimensions which are not the dimensions of mass.