Question

Draw a diagram to show conservation of mechanical energy in case of a free-falling body.

Accepted Answer

Let us suppose an object of mass $m$ is falling from a height $h$ to the ground.

Case A: The potential energy of the object at height $h$ on point A will be given by

$P E = m g h$

The kinetic energy of the object at height $h$ when the object is at point A will be

$K E = 0$

Therefore, the total mechanical energy, $M E = P E + K E = m g h + 0 = m g h$

Case B: Now, suppose the object has fallen to a certain distance of $x$ from that height and reached point B. Then the potential energy of the object at height $(h - x)$ from the ground is given by

$P E = m g (h - x)$

Using the third equation of motion, $v^{2} - u^{2} = 2 \times a \times s$

Since the initial velocity is zero. So the final velocity at point B is given by

$v^{2} = 2 \times g \times x$

The kinetic energy of the object at height $(h - x)$ is given by

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times m \times (2 \times g \times x) = m g x$

Therefore, the total mechanical energy at point B is given by

$M E = P E + K E = m g (h - x) + m g x = m g h$

Case C: Now, suppose the object has reached ground at point C. The potential energy of the object will be

$P E = 0$

Using the third equation of motion, when the initial velocity is zero and the object has travelled a distance of $h$ .

$v^{2} - u^{2} = 2 \times a \times s$

$v^{2} = 2 \times g \times h$

The kinetic energy of the object is given by

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times m \times (2 \times g \times h) = m g h$

Therefore, the total mechanical energy at point C be

$M E = P E + K E = 0 + m g h = m g h$

Hence, the mechanical energy in the case of a free-falling body is conserved.

Draw a diagram to show conservation of mechanical energy in case of a free-falling body.

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