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# Derive an expression of kinetic energy of a body of mass m and moving with velocity v, using dimensional analysis

Last updated date: 13th Jun 2024
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Hint: To obtain the expression of any physical derived quantity with respect to some other physical quantities then we need to know the dimension of those quantities. First we take the derived quantity and equate it to the quantities which it depends upon raised to some power times some constant. It is to be noted that the constant cannot be determined. Further we compare the powers of the equation on both the sides and determine the unknown power of the dependent quantities. This is how we will solve the above problem.

Let us first write the dimension of each of the quantities mentioned in the question.
Dimension of kinetic energy are $\left[ \text{M}{{\text{L}}^{2}}{{\text{T}}^{-2}} \right]$
Dimension of velocity are $\left[ {{\text{M}}^{0}}\text{L}{{\text{T}}^{-1}} \right]$
Dimensions of mass of a body is $\left[ \text{M}{{\text{L}}^{0}}{{\text{T}}^{0}} \right]$

It is given to us that the kinetic energy is dependent on mass and the velocity of a body. Hence we can say that $K.E=k{{v}^{a}}{{m}^{b}}....(1)$where k,a and b are all constants

Now let us replace the quantities in the above expression with their dimensions
\begin{align} & K.E=k{{v}^{a}}{{m}^{b}} \\ &\left[\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}\right]\text{=k}{{\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{-1}}} \right]}^{\text{a}}}{{\left[ \text{M}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}} \right]}^{\text{b}}} \end{align}

By using law of exponents i.e. ${{\left( {{\text{A}}^{\text{m}}} \right)}^{\text{n}}}\text{=}{{\text{A}}^{\text{mn}}}$ we get,

$\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}} \right]\text{=k}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{a}}}{{\text{T}}^{\text{-a}}} \right]\left[ {{\text{M}}^{\text{b}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}} \right]$
Again we will use one more law of exponents i.e. ${{\text{A}}^{\text{m}}}{{\text{A}}^{\text{n}}}\text{=}{{\text{A}}^{\text{m+n}}}$
$\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}} \right]\text{=k}\left[ {{\text{M}}^{b}}{{\text{L}}^{\text{a}}}{{\text{T}}^{\text{-a}}} \right]$

After comparing the powers of the same fundamental quantities we get,
b= 1 and a=2.
Now let us substitute the values in equation 1 and we get,

$\text{K}\text{.E=k}{{\text{v}}^{\text{2}}}\text{m}$
Hence,
$\text{K}\text{.E=k m}{{\text{v}}^{\text{2}}}$

From the above expression we can conclude that kinetic energy of a body in motion is directly proportional to its mass and directly proportional to the square of its velocity.

Note:
It is to be noted that the value of k is experimentally found to be half. The above expression is for a particle in motion. T any instant its kinetic energy is given by the above obtained expression. This method is actually useful to relate quantities when we cannot come to a final expression between the physical quantities.