three uses of dimensional analysis,kinetic energy,potential energy and work energy theorm
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Explanation:
The dimension of energy [E]=[M
1
L
2
T
−2
]
The dimension of mass [M]=[M
1
L
0
T
0
]
Dimension of velocity[V]=[M
0
L
1
T
−1
]
Let [E]=k[M]
x
[V]
y
Where k is the proportionality constant which is a dimensionless quantity.
Therefore,
[M
1
L
2
T
−2
]=k[M
1
L
0
T
0
]
x
[M
0
L
1
T
−1
]
y
So, x=1,y=2
Thus, We have
E=kmv
2
It is found that k=
2
1
Hence, E=
2
1
mv
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