Show that 

(a) KE = `\frac {1}{2}` mv²

and 

(b) PE𝖌 = mgh 

are dimensionally correct. 


Given:

(a) KE = `\frac {1}{2}` mv²
(b) PE𝖌 = mgh 


To Find:

(a)  To show KE = `\frac {1}{2}` mv²  is dimensionally correct.
(b)  To show PE𝖌 = mgh  is dimensionally correct?


Solution:

(a) is KE = `\frac {1}{2}` mv²  dimensionally correct?

Dimensions of L.H.S. of the equation = [KE] = [M L² T⁻²]

Dimensions of R.H.S. of the equation = [m] ｘ [v]²
                                                        = [M] ｘ [LT⁻¹]²
                                                        = [M] ｘ [LT⁻¹]ｘ [LT⁻¹]
                                                        = [ML²T⁻²]

L.H.S = R.H.S

Hence proved that the equation  KE = `\frac {1}{2}` mv²  is dimensionally correct.


(b)  To show PE𝖌 = mgh  is dimensionally correct.

Dimensions of L.H.S. of the equation = [PE𝖌] = [M L² T⁻²]

Dimensions of R.H.S. of the equation = [m] ｘ [g] ｘ [h]
                                                        = [M] ｘ [LT⁻²]ｘ [L]
                                                        = [ML²T⁻²]

L.H.S = R.H.S

Hence proved that the equation  PE𝖌 = mgh is dimensionally correct.

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