Question

A body is falling from a height h. After it has fallen a height $\dfrac{h}{2}$, it will possess:
A) Only potential energy.
B) Only kinetic energy.
C) Half potential energy and half kinetic energy.
D) More kinetic and less potential energy.

Answer 1

Hint: Kinetic energy is the energy developed due to motion of the object whereas the potential energy is the energy of the object which is possessed due to the position of the body. When an object is freely falling from height then the potential energy gets converted into kinetic energy.

Formula used:
The formula of the third relation of the Newton’s law of motion is given by,
$ \Rightarrow {v^2} - {u^2} = 2as$
Where final velocity is v the initial velocity is u the displacement is s and the acceleration of the body is a.
The formula of potential energy is given by,
$ \Rightarrow P.E = mgh$
Where mass of the body is m the acceleration is g and the height is h.
The formula of the kinetic energy is given by,
$ \Rightarrow K.E = \dfrac{1}{2}m{v^2}$
Where mass is m the velocity of the body is v.

Complete step by step solution:
It is given in the problem that a body is falling from height h and we need to tell what is the portion of kinetic and potential energy if the body has crossed half distance.
It is given in the problem that the body covers half distance therefore the potential energy of the motion is equal to,
The formula of potential energy is given by,
$ \Rightarrow P.E = mgh$
Where mass of the body is m the acceleration is g and the height is h.
As the distance travelled is equal to half of the total therefore,
$ \Rightarrow P.E = mgh$
$ \Rightarrow P.E = \dfrac{{mgh}}{2}$………eq. (1)
Let’s calculate the final velocity of the falling body,
The formula of the third relation of the Newton’s law of motion is given by,
$ \Rightarrow {v^2} - {u^2} = 2as$
Where final velocity is v the initial velocity is u the displacement is s and the acceleration of the body is a.
As the body was in rest so the initial velocity of the body is zero, the distance covered by the body is half of the total distance and the acceleration acting on the body is acceleration due to gravity therefore we get,
$ \Rightarrow {v^2} - {u^2} = 2as$
$ \Rightarrow {v^2} = 2 \cdot g \cdot \left( {\dfrac{h}{2}} \right)$
$ \Rightarrow {v^2} = g \cdot h$………eq. (2)
The formula of the kinetic energy is given by,
$ \Rightarrow K.E = \dfrac{1}{2}m{v^2}$………eq. (3)
Where mass is m the velocity of the body is v.
As the velocity of the body can be replaced from equation (2) into equation (3). Therefore the acceleration of the body is equal to,
$ \Rightarrow K.E = \dfrac{1}{2}m{v^2}$
$ \Rightarrow K.E = \dfrac{1}{2}m\left( {g \cdot h} \right)$
$ \Rightarrow K.E = \dfrac{1}{2}m \cdot g \cdot h$………eq. (4)
Comparing the equation (1) and equation (4) can observe that kinetic energy and potential energy are equal and half of the potential energy.
So the energy at the point after crossing half the distance is half kinetic energy and half potential energy.

The correct answer for this problem is option (C).

Note: It is advisable for students to remember and understand the formula of Newton's law of motion. If an object is freely falling from the height h then the acceleration acting on the body is acceleration due to gravity therefore there is conversion of potential energy into kinetic energy.