Question

total mechanical energy E of an object of mass m, at a height h above the ground is given as E = mgh+p^2/2m. Find the dimensions of p. (g = acceleration due to gravity)

Answer 1

Hope the images helps

Answer 2

The dimensions of p is $\bold{\left[\mathbf{M}^{1} \mathbf{L}^{1} \mathbf{T}^{-\mathbf{1}}\right]}$

Explanation:

The unit of mechanical energy E is Joule which is derived unit.

So,

$1 \ J =1 \mathrm{kg} \times \mathrm{m}^{2} \times \mathrm{s}^{-2}$

The dimensional constant for kg is M, m is L and s is T.

Thus,

$\text{Dimensional constant of }\mathrm{J}=\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$

The given equation is,

$\mathrm{E}=\mathrm{mgh}+\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

The S.I units and its corresponding dimensional constants for E, m, g and h are given below:

$\mathrm{E}=\mathrm{J}=\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$

$\mathrm{m}=\mathrm{Kg}=\left[\mathrm{M}^{1}\right]$

$\mathrm{g}=\mathrm{ms}^{-2}=\left[\mathrm{L}^{1} \mathrm{T}^{-2}\right]$

$\mathrm{h}=\mathrm{m}=\left[\mathrm{L}^{1}\right]$

mgh can be ignored, as the dimensional constant of mgh and $\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$ will be same as addition of two values can occur if their dimensional constant are same. So, E will also have same dimensional constant.

Thus, we can write the below equation to find the dimensional constant of p.

$\mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

$\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]=\left\{\frac{\mathrm{p}^{2}}{\left[\mathrm{M}^{1}\right]}\right\}$

The value 2 is ignored as it does not affect the dimensional constants.

$\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}\right] \times\left[\mathrm{M}^{1}\right]=\mathrm{p}^{2}$

$\left[\mathrm{M}^{2} \mathrm{L}^{2} \mathrm{T}^{-2}\right]=\mathrm{p}^{2}$

Taking square root on both sides,

$\mathrm{p}=\left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-1}\right]$