Question

Write dimensional formula of energy and acceleration.

Answer 1

Hint : To solve this question, we have to use the basic formulas of the energy and the acceleration. Then we have to write the dimensions of the quantities with which these are related. On simplification, we will get the required dimensional formula.

Complete step by step answer
We know that the acceleration is defined as the rate of change of velocity. So it is given by
$a = \dfrac{{\Delta v}}{{\Delta t}}$
Taking dimensions of both the sides, we have
$\left[ a \right] = \dfrac{{\left[ {\Delta v} \right]}}{{\left[ {\Delta t} \right]}}$ ………...(1)
Now, the velocity is defined as the rate of change of position of a particle, that is,
$v = \dfrac{{\Delta x}}{{\Delta t}}$
Taking the dimensions on both the sides, we get
$\left[ v \right] = \dfrac{{\left[ {\Delta x} \right]}}{{\left[ {\Delta t} \right]}}$ ………...(2)
We know that the dimension of the position is the same as that of length. So
$\left[ {\Delta x} \right] = \left[ {{M^0}{L^1}{T^0}} \right]$ ………...(3)
Also, the dimension of the change in time is given by
$\left[ {\Delta t} \right] = \left[ {{M^0}{L^0}{T^1}} \right]$ ………...(4)
Substituting (3) and (4) in (2) we get
$\left[ v \right] = \dfrac{{\left[ {{M^0}{L^1}{T^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}} \right]}}$
On solving we get the dimensions of velocity as
$\left[ v \right] = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
The dimensions of the change in velocity will be the same as that of the velocity, therefore
$\left[ {\Delta v} \right] = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$ ………...(5)
Substituting (4) and (5) in (1) we get
$\left[ a \right] = \dfrac{{\left[ {{M^0}{L^1}{T^{ - 1}}} \right]}}{{\left[ {{M^0}{L^0}{T^1}} \right]}}$
$\Rightarrow \left[ a \right] = \left[ {{M^0}{L^1}{T^{ - 2}}} \right]$ ………...(6)
Therefore, the dimension formula of the acceleration is $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$ .
Now, we know that the work done is a form of energy. The work done is defined as the product of the force and the displacement. So the energy can also be given by
$E = Fs$
From the Newton’s second law of motion, we know that
$F = ma$
Substituting this above, we get
$E = mas$
Writing the dimensions of both the sides, we get
$\left[ E \right] = \left[ m \right]\left[ a \right]\left[ s \right]$ ………...(7)
We know that the dimensions of mass are given by
$\left[ m \right] = \left[ {{M^1}{L^0}{T^0}} \right]$ ………...(8)
Also, we know that the dimensions of displacement are given by
$\left[ s \right] = \left[ {{M^0}{L^1}{T^0}} \right]$ ………...(9)
Substituting (6) (8) and (9) in (7) we get
$\left[ E \right] = \left[ {{M^1}{L^0}{T^0}} \right]\left[ {{M^0}{L^1}{T^{ - 2}}} \right]\left[ {{M^0}{L^1}{T^0}} \right]$
On simplifying we get
$\left[ E \right] = \left[ {{M^0}{L^2}{T^{ - 2}}} \right]$
Hence, the dimensional formula of the energy is $\left[ {{M^0}{L^2}{T^{ - 2}}} \right]$ .

Note
For solving such types of questions, it is required to obtain each physical quantity given in the problem in terms of the fundamental or base quantities. So we must know the physical formula which can relate the quantity with the fundamental quantities. It must be noted that there can be multiple such formulas and any one of them can be used for this purpose.