Answer:
[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}[/tex]
Explanation:
[tex] \underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}[/tex]
Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :
[tex] \sf{W = FD}[/tex]
⇒[tex] \sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)[/tex]
⇒[tex] \sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}[/tex]
⇒[tex] \sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }[/tex]
⇒[tex] \sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}[/tex]
The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²
∴ [tex] \sf{KE= \frac{1}{2} m {v}^{2} }[/tex]
[tex] \sf{ \underline{ \bold{ {proved}}}}[/tex]
Hope I helped!
Best regards!!
