Question

Check the dimensions of the equation 1÷2mv2 -1÷2mu2=Fs in physics​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

From LHS -

$\sf{\dfrac{1}{2}mv^{2} - \dfrac{1}{2}mu^{2}}$

$\sf{\dfrac{1}{2}m (v^{2} - u^{2})}$

By Using Third equation of Newton's motion

We know -

$\sf{v^{2} - u^{2} = 2as}$

Then ,

$\sf{= \dfrac{m}{2}(2as)}$

$\sf{= mas}$

By Second law of Newton's motion -

$\sf{F = ma}$

then ,

$\sf{= Fs}$

Now , from Dimensions -

$\sf{Dimensions \: of \: Mass = [M^{1}L^{0}T^{0}]}$

$\sf{Dimension \: of \: acceleration =[M^{0}L^{1}T^{-2}]}$

$\sf{Dimension \: of \: distance =[M^{0}L^{1}T^{0}]}$

$\sf{Dimension \: of \: force =[M^{1}L^{1}T^{-2}]}$

$\sf\boxed{mas = Fs}$

$\sf{[M^{1}][M^{0}L^{1}T^{-2}][M^{0}L^{1}T^{0}] =[M^{1}L^{1}T^{-2}][M^{0}L^{1}T^{0}]}$

$\sf{[M^{1+0+0} \: L^{0+1+1} \: T^{0-2+0}]=[M^{1+0} \: L^{1+1} \: T^{-2+0}]}$

$\sf{[M^{1}L^{2}T^{-2}] = [M^{1}L^{2}T^{-2}]}$

$\sf{L . H . S = R . H . S}$