Question

Q. States the law of conservation of energy with its derivation. Image or write the answer.

Answer 1

Law of Conservation of Energy

According to this law energy is conserved at every point but never created or destroyed.

Here, mechanical energy = K.E + P.E

Let object fall from point A to C with point B in middle.

At point A, K.E is always 0.

Mechanical energy = 0 + mgh = mgh

At point B, distance for K.E is v^2 = u^2 + 2as

When u = 0

Then v^2 = 2gx

Mechanical energy = 1/2 × 2gx × m + mg( h - x )

= mgx + mgh - mgx = mgh

At point C no P.E i.e 0,

Mechanical energy = 1/2× 2gh × m + 0 = mgh

Hence the energy is conserved

Answer 2

THE LAW OF CONSERVATION OF ENERGY :

the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

DERIVATION

Let us consider a body of mass "m" placed at a point A. "h" is the height of the body above the ground. "s" is the distance, acceleration is "g", velocity of the body at C is "v1" and "v" is the velocity of the body at B.

• Velocity at the point A =0
• (i.e)u =0

Case 1: AT POINT "A"

P. E = mgh

K. E = 0

Total energy = K. E +P. E

= 0 +mgh

=mgh

Case 2 : AT POINT "C"

When the body moves from A to C. It covers a distance "s". If v1 is the velocity at C

${v1}^{2} + {u}^{2} = 2as \\ \\ {v1}^{2} + 0 = 2as \\ \\ {v1}^{2} = 2as$

K. E

$= \frac{1}{2} m {v1}^{2} \\ \\ but \: \: {v1}^{2} = 2gs$

THEREFORE

K. E

$= \frac{1}{2} m2gs \\ \\ = mgs$

P. E

$= mg(h - s) \\ \\ = mgh - mgs$

T. E

$= mgh - mgs + mgs \\ \\ = mgh$

Case 3: AT POINT "B"

${v}^{2} - {u}^{2} = 2as \\ \\ {v}^{2} = 2gh$

K. E

$= \frac{1}{2} m {v}^{2} \\ \\ ( {v}^{2} = 2gh) \\ \\ = \frac{1}{2} m2gh \\ \\ = mgh$

P. E = 0

T. E = K. E +P. E

= 0 + mgh

=mgh

FROM THE ABOVE DERIVATION WE CAN SAY THAT THE TOTAL ENERGY OF THE BODY THROUGHOUT THE FREE FALL IS CONSERVED.