Question

show that total machinical energy of a body follow freely under gravity always ramain constant

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Show :- }}}$

$\textsf{ \to The Total Mechanical Energy of a body remains} \\\textsf{\qquad constant under free fall. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Proof:- }}}$

Let us consider a free falling ball , falling from a tower of height h . Let a point C lie somewhere between A and B . Let us take BC = x , hence AC = ( h - x ). Let's Calculate the total mechanical energies at points A , B and C . Also we know that the total mechanical energy is equal to the sum of Potential energy ( $\sf E_{(potential)}$ ) and Kinetic energy ( $\sf E_{(Kinetic)}$ )

$\large\underline\purple{\textsf{ T.M.A at point A :- } }$

$\sf:\implies \pink{T.M.A_A = P.E._A + K.E._A} \\\\\sf:\implies T.M.A_A = (m)(g)(h)+ \dfrac{1}{2}(m)(v)^2 \\\\\sf:\implies T.M.A_A = (m)(g)(h)+ \dfrac{1}{2}(m)(0m/s)^2\\\\\sf:\implies T.M.A._A= mgh + 0 \\\\\sf:\implies \boxed{\sf T.M.A_A= mgh }$

$\rule{200}2$

$\large\underline\purple{\textsf{ T.M.A at point B :- } }$

$\sf:\implies \pink{T.M.A_B = P.E._B + K.E._B} \\\\\sf:\implies T.M.A_B = (m)(g)(x)+ \dfrac{1}{2}(m)(v)^2 \\\\\sf:\implies T.M.A_B = (m)(g)(x)+ \dfrac{1}{2}(m)(\sqrt{2g(h-x)})^2\\\\\sf:\implies T.M.A._B= mgx + \dfrac{1}{2}\times (m) \times 2g(h-x)\\\\\sf:\implies T.M.A._B = mgx + mg(h-x)\\\\\sf:\implies T.M.A.= mgx + mgh - mgx\\\\\sf:\implies \boxed{\sf T.M.A_B= mgh }$

$\rule{200}2$

$\large\underline\purple{\textsf{ T.M.A at point C :- } }$

$\sf:\implies T.M.A_C = P.E._C + K.E._C \\\\\sf:\implies T.M.A_C = (m)(g)(0) + \dfrac{1}{2}\times (m)\times (\sqrt{2gh})^2 \\\\\sf:\implies T.M.A_C = 0 + \dfrac{1}{2}\times m \times 2gh \\\\\sf:\implies \boxed{\sf T.M.A_C= mgh }$

Hence here we can see that the TMA at point A is equal to the TMA at point B is equal to the TMA energy at point C .

$\underset{\red{\bf Hence \ Proved}}{\underbrace{{\boxed{\pink{\mathbb{ T.M.A._A = T.M.A._B = T.M.A._C }}}}}}$

Answer 2

Total mechanical energy = PE + KE

Consider a particle is in free fall from rest and is at height (h) also let it's mass be (m). And acceleration is (g).

At the top most point:-

PE = m·g·h

KE = ½·m·v² = ½·m·0² = 0

∴ Total mechanical energy = PE + KE = mgh + 0 = mgh

At the middle of the path:-

PE = m·g·h/2

KE = ½·m·v′²

By v² = u² + 2g∆x,

⇒ v′² = 0² + gh/2

(x = h)

So, KE = ½·m·g·h = mgh/2

∴ Total mechanical energy = PE + KE = mgh/2 + mgh/2 = mgh

At the end:-

PE = mgh = m·g·0 = 0

KE = ½·m·v″²

By v² = u² + 2g∆x,

⇒ v² = 2gh

So, KE = ½·m·2gh = mgh

∴ T.M.E. = KE + PE = mgh + 0 = mgh

So, in all cases, energy remains conserved.