Question

The total mechanical energy $E$ possessed by a body of mass$'m'$ , moving with a velocity $'v'$ at a height $'h'$ is given by: $E = \dfrac{1}{2}m{v^2} + mgh$. Make $'m'$ the subject of the formula.
$(A)m = \dfrac{{2E}}{{{v^2} + gh}}$
$(B)m = \dfrac{{2E}}{{{v^2} + 2gh}}$
$(C)m = \dfrac{{3E}}{{{v^2} + 2gh}}$
$(D)m = \dfrac{E}{{{v^2} + 2gh}}$

Answer 1

Hint:In the problem, the formula of the total energy of a body is given. The total energy is the sum of kinetic energy and potential energy. The mass has to make the subject of the formula. It means, the formula of energy has to be simplified and then represent the mass in terms of energy and other given quantities. The answer can be checked by putting the units of the quantities on the right-hand side and see if the quantities give the unit of mass after simplification.

 Complete step by step solution:
The total mechanical energy $E$is given by the sum of the kinetic energy and potential energy.
$E = \dfrac{1}{2}m{v^2} + mgh$
$'m'$is the mass, $'v'$is the velocity, $'h'$is the height, and $'g'$ is the acceleration due to gravity.
We have to simplify the equation to make the mass as the subject of the formula.
So, $E = \dfrac{1}{2}m{v^2} + mgh$
$ \Rightarrow 2E = m({v^2} + gh$
$ \Rightarrow m({v^2} + gh) = 2E$
$ \Rightarrow m = \dfrac{{2E}}{{({v^2} + gh)}}$
So, the answer is $m = \dfrac{{2E}}{{({v^2} + gh)}}$
Hence, the right answer is in Option (A).

Note:The answer is right or wrong; this can be checked by using the units of the quantities. let us take the SI units of them.
The SI unit of energy is Joule. Joule can be written as kilogram-meter/ second2 ( $kg.{m^2}/{s^2}$).
 The SI unit of velocity is meter/second ($m/s$), the Si Unit of acceleration is meter/second2 $(m/{s^2})$ , and the SI unit of the height is meter ($m$ ).
Now, if we put the units in the equation - $m = \dfrac{{2E}}{{({v^2} + gh)}}$, we will get the SI unit of mass.
\[\therefore {\text{unit of mass = }}\dfrac{{2\dfrac{{kg.{m^2}}}{{{s^2}}}}}{{{{\left( {\dfrac{m}{s}} \right)}^2} + \left( {\dfrac{m}{{{s^2}}} \times m} \right)}}\]
\[ \Rightarrow {\text{unit of mass = }}\dfrac{{2\dfrac{{kg.{m^2}}}{{{s^2}}}}}{{2\dfrac{{{m^2}}}{{{s^2}}}}}\]
\[ \Rightarrow {\text{unit of mass = }}kg\]
As we know the SI unit of the mass is kg. hence the answer is correct.