This relationship was first published by Sir Issac Newton. His law of universal gravitation says that the force (F) of gravitational attraction between two objects with Mass1 and Mass2 at distance D is: F = G(mass1*mass2)/D squared

4 0

1 year ago

Show that (a)KE=1/2mv2

evablogger [386]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}$

Explanation:

$\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}$

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 :

$\sf{W = FD}$

⇒ $\sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)$

⇒ $\sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}$

⇒ $\sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }$

⇒ $\sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}$

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

∴ $\sf{KE= \frac{1}{2} m {v}^{2} }$

$\sf{ \underline{ \bold{ {proved}}}}$

Hope I helped!

Best regards!!

5 0

2 years ago