Question

A body is falling from a height $\mathrm{h}$. After it has fallen from a height $\frac{\mathrm{h}}{2}$, it will possess

• A. only potential energy
• B. only kinetic energy
• C. half potential and half kinetic energy
• D. more kinetic and less potential energy

Answer 1

The correct option is C

half potential and half kinetic energy

Step 1. Given data

It is given that a body is falling from a height $\mathrm{h}$. After it has fallen from a height $\frac{\mathrm{h}}{2}$.

Step 2. Formula to be used

The energy possessed by the body due to virtue of its motion is called Kinetic energy.

The kinetic energy is,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2$

Here, $\mathrm{m}$ is mass, $\mathrm{v}$ is velocity and $\mathrm{K}.\mathrm{E}.$ is kinetic energy.

The energy possessed by a body due to the virtue of its position or configuration is called potential energy.

$\mathrm{P}.\mathrm{E}.=\mathrm{mgh}$

Here, $\mathrm{m}$ is mass, $\mathrm{g}$ is gravitational field strength and $\mathrm{h}$ is height.

Step 3. Find the energy which possesses by the given height

Let a body of mass ′m′ placed at a height $\prime \mathrm{h}\prime$ above the ground, and start falling down from rest which is shown in above figure.
When the body falls from a height $\prime \mathrm{h}\prime$ to $\frac{\mathrm{h}}{2}$, the potential energy is given as $\frac{\mathrm{mgh}}{2}$.
The body covers the distance $\frac{\mathrm{h}}{2}$ with a velocity $\mathrm{v}$. We make use of the third equation of motion to obtain the velocity of the body.

According to the third equation of motion is,

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

Here, $\mathrm{v}$ is the velocity at any point in time, $\mathrm{u}$ is the initial velocity, $\mathrm{a}$ is the acceleration and $\mathrm{s}$ is the distance travelled. From the given,

$\mathrm{u}=0$

$\mathrm{a}=\mathrm{g}$

$\mathrm{s}=\mathrm{x}$

So,

${\mathrm{v}}^2-0=\frac{2\mathrm{gh}}{2}$

${\mathrm{v}}^2=\mathrm{gh}$

Now,

Kinetic energy $=\frac{1}{2}{\mathrm{mv}}^2$

$=\frac{\mathrm{mgh}}{2}$

So, at height $\frac{\mathrm{h}}{2}$ the energy possessed is half potential energy and half the kinetic energy.

Hence, option $\mathrm{C}$ is the correct answer.