Question

Derive the expression for K.E of body having mass m moving with the speed of V meter per sec.

Answer 1

Let ,

$\star$A body of mass = m

$\star$Initial position of body is at rest

$\star$An applied force on a body = F

$\star$It travels a distance = s

So , Work done (W) = F × s

By newton's second law of motion , F = ma

$\implies \sf\frac{W}{s} = ma\\ \\ \sf \implies W= ma × s \: \: - - - - \: ( \sf i)$

Where , a = acceleration

By using second equation of motion v² - u² = 2as , we get

$\implies \sf \frac{( {v}^{2} - {u}^{2} )}{2} = as$

Put the value of as = (v² - u²)/2 in $\sf (i)$

$\sf \implies W = \frac{m \times ( {v}^{2} - {u}^{2} )}{2}$

The intial velocity of an body is zero { Given }

$\sf \implies W = \frac{m \times ( {v}^{2} - {(0)}^{2} )}{2} \\ \\ \sf \implies W = \frac{1}{2} \times m {(v)}^{2}$

This work done is stored as kinetic energy of the object.

So ,

$\large \fbox{\fbox{ \sf \: \: Kinetic \: energy = \frac{1}{2} \times m {(v)}^{2}} \: \: }$