Question

Show the total energy remain constant in a gravitational field.[4] koi bhi answer kar diy na me use thanks kar dungi

Answer 1

Answer:

this is the answer hope it helps

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \underline{{ \huge{ \mathfrak{ \: Q}}} \mathfrak{uestion :-}}}$

Show the total energy remain constant in gravitational field.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \underline{{ \huge{ \mathfrak{ \: T}}} \mathfrak{o \: \: prove :- }}}$

The total energy = constant

K.E. + P.E. = constant

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \underline{{ \huge{ \mathfrak{ \:S}}} \mathfrak{olution :- }}}$

(Also refer the attachment given)

In the above give fig.

Let

• Mass of the body = m
• Height = h
• Under gravity = g

At position A (at height h above the ground) :

Potential Energy = mgh

Kinetic energy = 0 (as it is above the ground)

∴ Total energy = Potential energy + kinetic energy

⇰ mgh + 0

⇰ mgh ...(i)

At position B (when it has fallen through to a distance x) :

Let:

• $\rm \: velocity = v_1$
• initial velocity = u
• distance (S) = x
• a = g

We know that,

$\rm ⇰ \: v² = u² + 2aS$

$\rm ⇰ \: v_1² = 0 + 2gx$

$\rm⇰ \: v_1² = 2gx$

$\rm∴ \: K.E. = \dfrac{1}{2} mv_1²$

$\rm⇰ \: \dfrac{1}{2}m(2gx)$

$\rm⇰ \: \dfrac{1}{ \cancel2}m \times \cancel{2}gx$

$\rm \rm⇰ \: mgx$

Now,

for potential energy at B,

h = h - x

∴ P.E. = mgh

⇰ mg(h-x)

∴ Total energy = P.E. + K.E.

⇰ mg(h-x) + mgx

⇰ mgh - mgx + mgx

⇰ mgh ...(ii)

At position C (on the ground) :

P.E. = 0

Let:

• velocity = v
• initial velocity = u
• distance (S) = h
• a = g

We know that,

⇰ v² = u² + 2aS

⇰ v² = 0 + 2gh

⇰ v² = 2gh

$\rm ∴ \: K.E. = \dfrac{1}{2} mv²$

$\rm⇰ \: \dfrac{1}{2} m(2gh)$

$\rm \: \rm⇰ \: \dfrac{1}{ \cancel2} m \times \cancel{ 2}gh$

$\rm⇰ \: mgh$

∴ Total energy = P.E. + K.E.

⇰ 0 + mgh

⇰ mgh ...(iii)

Now,

From eq. (i) , (ii) & (iii)..

We can conclude that,

⇰ Sum of Total energy = mgh

⇰ Total energy = constant... (proved)