Question

Check the correctness of the equation,
$FS=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}$

Answer 1

The correctness of any equation can be checked using the dimensional analysis. The dimensions of the quantities should be the same on both sides of the equality for an equation to be correct. Derive each quantity in terms of the fundamental quantities to find the dimensional formula of any quantity.
Formula used:
$\text{Force}=\text{mass}\times \text{acceleration}$

Complete answer:
Dimensions are defined as the fundamental quantities which cannot be derived from any other quantity. Every other quantity other than the fundamental quantities can be derived using the fundamental quantities. Some of the fundamental quantities include Mass $\left( \text{M} \right)$, Length $\left( \text{L} \right)$ and Time $\left[ \text{T} \right]$.
Let us look at the given equation.
$FS=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}$
Here, $F$ represents force, $S$ is displacement, $m$ is mass and $v$ and $u$ are velocities.
Let us determine the dimensions of the quantities on the left side of equality. The dimensions of force can be determined as,

& \left[ F \right]=\text{mass}\times \text{acceleration} \\\ & =\text{mass}\times \dfrac{\text{velocity}}{\text{time}} \\\ & =\text{mass}\times \dfrac{\dfrac{\text{displacement}}{\text{time}}}{\text{time}} \\\ & =\text{M}\times \dfrac{\dfrac{\text{L}}{\text{T}}}{\text{T}} \\\ & =\text{ML}{{\text{T}}^{-2}} \end{aligned}$$ The dimensions of displacement are $\left[ \text{L} \right]$. Thus, the dimensions of the quantities on left hand side can be written as, $\begin{aligned} & LHS=\left[ \text{ML}{{\text{T}}^{-2}} \right]\cdot \left[ \text{L} \right] \\\ & =\left[ \text{M}{{\text{L}}^{2}}{{\text{T}}^{-2}} \right] \end{aligned}$ Now let us determine the dimensions of the quantities on the right hand side. $\begin{aligned} & \left[ v \right]=\dfrac{\text{displacement}}{\text{time}} \\\ & =\dfrac{\text{L}}{\text{T}} \\\ & =\text{L}{{\text{T}}^{-1}} \end{aligned}$ The dimensions of the quantities on the right-hand side are determined as, $$\begin{aligned} & RHS=\left[ \text{M} \right]{{\left[ \text{L}{{\text{T}}^{-1}} \right]}^{2}}-\left[ \text{M} \right]{{\left[ \text{L}{{\text{T}}^{-1}} \right]}^{2}} \\\ & =\text{M}{{\text{L}}^{2}}{{\text{T}}^{-2}} \end{aligned}$$ See that the numerical coefficient does not have any dimensions and are ignored while doing the dimensional analysis. Also, one thing that needs to be mentioned is that mathematical operations such as addition and subtraction can only be performed if the dimensions of both the quantities are the same and the result would have the same dimensions. Now, we can see that the dimensions on the left-hand side and right-hand side are equal. Thus, the equation is correct. **Note:** Every quantity has to be derived in terms of the fundamental quantities for the dimensional analysis and to check the correctness of an equation. Mathematical operations such as addition and subtraction can be performed on the quantities only if the dimensions of both the quantities are same and the result has the same dimensions. For example, dimensions of the difference of two lengths would also be length.