Question

Prove that total mechanical energy of a free falling object remains always constant.


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Answer 1

Answer:

• The total Mechanical Energy remains Constant.

Statement:

Law of conservation of energy:

Energy can neither be created nor destroyed but it can be transformed to one form to other form.

Explanation:

$\rule{300}{1.5}$

Case-1

At this time the body is at Position " A " ,

Where,

• Its Velocity (v) is Zero.
• Its at a Height " h " from the ground.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × h

⇒ P.E = m g h J

Now,

⇒ K.E = 1 / 2 m v²

Substituting the values,

⇒ K.E = 1 / 2 × m × (0)²

⇒ K.E = 1 / 2 × 0

⇒ K.E = 0 J

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = 0 + m g h

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'A' is m g h Joules.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Case-2

At this time the body is at Position " B " ,

Where,

• Its Velocity (v) is v₁ .
• Its at a Height " (h - s) " from the ground.
• And, has travelled a Distance "s" from top.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × ( h - s )

⇒ P.E = m g ( h - s ) J

Now,

⇒ K.E = 1 / 2 m v²

Firstly finding the velocity of the particle,

Applying third Kinematic equation.

⇒ v² - u² = 2 a s

Substituting,

⇒ (v₁)² - 0 = 2 × g × s

⇒ v₁² = 2 g s

⇒ v₁ = √ 2 g s

Substituting the value of velocity in Kinetic Energy.

⇒ K.E = 1 / 2 × m × (v₁)²

⇒ K.E = 1 / 2 × m × (√ 2 g s)²

⇒ K.E = 1 / 2 × m × 2 g s

⇒ K.E = m × g × s

⇒ K.E = m g s

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = m g s + m g ( h - s )

⇒ T.E = m g s + m g h - m g s

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'B' is m g h Joules.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Case-3

At this time the body is at Position " C " ,

Where,

• Its Velocity (v) is v .
• Its at a Height " 0 m " from the ground.
• And, has travelled a Distance "h" from top.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × 0

⇒ P.E = 0 J

Now,

⇒ K.E = 1 / 2 m v²

Firstly finding the velocity of the particle,

Applying third Kinematic equation.

⇒ v² - u² = 2 a s

Substituting,

⇒ (v)² - 0 = 2 × g × h

⇒ v² = 2 g h

⇒ v = √ 2 g h

Substituting the value of velocity in Kinetic Energy.

⇒ K.E = 1 / 2 × m × (v)²

⇒ K.E = 1 / 2 × m × (√ 2 g h)²

⇒ K.E = 1 / 2 × m × 2 g h

⇒ K.E = m × g × h

⇒ K.E = m g h

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = m g h + 0

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'C' is m g h Joules.

Hence Proved!

# Refer the Attachment for figure.

Assumptions:

• " m " is the mass of the body.
• " g " is the acceleration due to gravity.

$\rule{300}{1.5}$

Answer 2

Answer:

The total Mechanical Energy remains Constant.

Statement:

Law of conservation of energy:

Energy can neither be created nor destroyed but it can be transformed to one form to other form.

Explanation:

$$\rule{300}{1.5}$$

Case-1

At this time the body is at Position " A " ,

Where,

Its Velocity (v) is Zero.

Its at a Height " h " from the ground.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × h

⇒ P.E = m g h J

Now,

⇒ K.E = 1 / 2 m v²

Substituting the values,

⇒ K.E = 1 / 2 × m × (0)²

⇒ K.E = 1 / 2 × 0

⇒ K.E = 0 J

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = 0 + m g h

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'A' is m g h Joules.

$$\rule{300}{1.5}$$

$$\rule{300}{1.5}$$

Case-2

At this time the body is at Position " B " ,

Where,

Its Velocity (v) is v₁ .

Its at a Height " (h - s) " from the ground.

And, has travelled a Distance "s" from top.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × ( h - s )

⇒ P.E = m g ( h - s ) J

Now,

⇒ K.E = 1 / 2 m v²

Firstly finding the velocity of the particle,

Applying third Kinematic equation.

⇒ v² - u² = 2 a s

Substituting,

⇒ (v₁)² - 0 = 2 × g × s

⇒ v₁² = 2 g s

⇒ v₁ = √ 2 g s

Substituting the value of velocity in Kinetic Energy.

⇒ K.E = 1 / 2 × m × (v₁)²

⇒ K.E = 1 / 2 × m × (√ 2 g s)²

⇒ K.E = 1 / 2 × m × 2 g s

⇒ K.E = m × g × s

⇒ K.E = m g s

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = m g s + m g ( h - s )

⇒ T.E = m g s + m g h - m g s

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'B' is m g h Joules.

$$\rule{300}{1.5}$$

$$\rule{300}{1.5}$$

Case-3

At this time the body is at Position " C " ,

Where,

Its Velocity (v) is v .

Its at a Height " 0 m " from the ground.

And, has travelled a Distance "h" from top.

Now,

⇒ P.E = m g h

Substituting the values,

⇒ P.E = m × g × 0

⇒ P.E = 0 J

Now,

⇒ K.E = 1 / 2 m v²

Firstly finding the velocity of the particle,

Applying third Kinematic equation.

⇒ v² - u² = 2 a s

Substituting,

⇒ (v)² - 0 = 2 × g × h

⇒ v² = 2 g h

⇒ v = √ 2 g h

Substituting the value of velocity in Kinetic Energy.

⇒ K.E = 1 / 2 × m × (v)²

⇒ K.E = 1 / 2 × m × (√ 2 g h)²

⇒ K.E = 1 / 2 × m × 2 g h

⇒ K.E = m × g × h

⇒ K.E = m g h

Therefore,

⇒ T.E = K.E + P.E

Substituting the values,

⇒ T.E = m g h + 0

⇒ T.E = m g h J

∴ The Total Energy (T.E) of the body at position 'C' is m g h Joules.

Hence Proved!

# Refer the Attachment for figure.

Assumptions:

" m " is the mass of the body.

" g " is the acceleration due to gravity.

$$\rule{300}{1.5}$$