Question

Define mechanical energy of a body. Show that mechanical energy is conserved when a body falls freely under gravity.

Accepted Answer

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic or potential.

Conservation of mechanical energy during free fall under gravity:

Consider and object mass $m$ at point $A$ at a height $h$ from ground level.

Initially body is at rest. It has only potential energy.

Its $K . E . = 0$

At $A$ total energy $= K . E . + P . E . = 0 + m g h$ $= m g h$ ..................(i)

Body starts falling and reaches $B$ . At $B$ it has both $K . E .$ and $P . E .$

Energy at $B = K . E . + P . E .$ ................(ii)

$K . E .$ at $B = \frac{1}{2} m {v^{2}}_{B}$ where $v_{B}$ is velocity acquired by object at $B$

${v^{2}}_{B} - 0 = 2 g x$ where is initial velocity at $A$ or ${v^{2}}_{B} = 2 g x$

$K . E .$ at $B = \frac{1}{2} m {v^{2}}_{B}$

$= \frac{1}{2} m (2 g x)$

$= m g x$ ..................(iii)

$P . E .$ at $B = m g (h - x)$ .............................(iv)

Where $(h - x)$ is height above ground level.

Total energy at $B = m g x + m g (h - x)$

$= m g h$ ................................(v)

When body reaches point $C$ it touches the ground.

$K . E .$ at $C = \frac{1}{2} m {v^{2}}_{C}$ where $v_{C}$ is velocity at point $C$

According to $v^{2} - u^{2} = 2 a s$

${v^{2}}_{C} - 0^{2} = 2 g h$ ..............(vi)

${v^{2}}_{C} = 2 g h$

Because $K . E . = \frac{1}{2} m \times 2 g h = m g h$

Kinetic energy $= m g h$ at $C$

Potential energy at $C = 0$

Because, total energy $= K . E . + P . E .$ $= m g h$

Mechanical energy is conserved.

G L Mittal and TARUN MITTAL Solutions for Chapter: Work, Power and Energy, Exercise 1: QUESTIONS

G L Mittal Physics Solutions for Exercise - G L Mittal and TARUN MITTAL Solutions for Chapter: Work, Power and Energy, Exercise 1: QUESTIONS

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