Total mechanical energy $E$ of an object of mass $m$ , at a height $h$ above the ground is given by the relation
$E = m g h + \frac{p^{2}}{2 m}$
Find the dimension of p. ( $g =$ acceleration due to gravity)

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Solution

It is given that

$E = m g h + \frac{p^{2}}{2 m}$

$2 m E = 2 m^{2} g h + p^{2}$

$p^{2} = 2 (2 m E - m^{2} g h)$

Writing dimensional formula of m,g,E and h

$m = [M]$

$g = [L T^{- 2}]$

$h = [L]$

$E = [M L^{2} T^{- 2}]$

$p^{2} = [M^{2} L^{2} T^{- 2}]$

$p = [M L T^{- 1}]$

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Total mechanical energy E of an object of mass m, at a height h above the ground is given by the relation E=mgh+p22mFind the dimension of p. (g=acceleration due to gravity)

It is given that E=mgh+p22m 2mE=2m2gh+p2 p2=2(2mE−m2gh) Writing dimensional formula of m,g,E and h m=[M] g=[LT−2] h=[L] E=[ML2T−2] p2=[M2L2T−2] p=[MLT−1]

Total mechanical energy $E$ of an object of mass $m$ , at a height $h$ above the ground is given by the relation
$E = m g h + \frac{p^{2}}{2 m}$
Find the dimension of p. ( $g =$ acceleration due to gravity)

It is given that

$E = m g h + \frac{p^{2}}{2 m}$

$2 m E = 2 m^{2} g h + p^{2}$

$p^{2} = 2 (2 m E - m^{2} g h)$

Writing dimensional formula of m,g,E and h

$m = [M]$

$g = [L T^{- 2}]$

$h = [L]$

$E = [M L^{2} T^{- 2}]$

$p^{2} = [M^{2} L^{2} T^{- 2}]$

$p = [M L T^{- 1}]$