Question

An object of mass \[40\,{\text{kg}}\] is raised to a height of \[5\,{\text{m}}\] above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Answer 1

Hint: First of all, we will find the potential energies of the object at the maximum height and half-way. We know, the sum of potential and kinetic energies at half-way gives the total kinetic energy at the maximum height. We will substitute the values and manipulate it accordingly.

Complete step by step solution:
Mass of the object \[ = 40\,{\text{kg}}\]
Acceleration due to gravity \[ = 9.8\,{\text{m}}{{\text{s}}^{ - 2}}\]
Height or position of the object from the ground \[ = 5\,{\text{m}}\]
We know, potential energy of an object which is at a certain height, is given by the formula:
\[P.E = m \times g \times h\] …… (1)
Where,
\[P.E\] indicates total potential energy of the object at the mentioned height.
\[m\] indicates mass of the object.
\[g\] indicates acceleration due to gravity.
\[h\] indicates height or position of the object from the ground.
Now substituting the required values in the equation (1), we get:
$P.E = m \times g \times h $
$\Rightarrow P.E = 40 \times 9.8 \times 5 $
$\Rightarrow P.E = 1960\,{\text{J}} $
Thus, the potential energy of the object which is at a height of \[5\,{\text{m}}\] is found out to be \[1960\,{\text{J}}\] .
The object has the maximum potential energy when it is at a height of \[5\,{\text{m}}\] . But when the object is allowed to fall such that it reaches the half of the previous height from the ground, in a situation when it is half way down, the new position of the object will be:
${h_1} = \dfrac{5}{2}\,{\text{m}} \\
\Rightarrow {h_1} = 2.5\,{\text{m}} $
Now the potential energy of the object when it is half way down, can be calculated by the same formula:
\[P.{E_1} = m \times g \times {h_1}\] …… (2)
Where,
\[P.{E_1}\] indicates new potential energy when it is half-way down.
\[{h_1}\] indicates the new position of the object.
Now, substituting the required values in equation (2), we get:
$P.E = m \times g \times h \\
\Rightarrow P.E = 40 \times 9.8 \times 2.5 \\
\Rightarrow P.E = 980\,{\text{J}} $
Thus, the potential energy of the object which when it is half-way down is found out to be \[980\,{\text{J}}\] .
According to the law of conservation of energy, energy can neither be created nor be destroyed, but can only be converted from one form to another.
The total potential energy of the object is the sum of the potential energy of the object at half-way down and the kinetic energy of the object at half-way down. Mathematically, we can write:
\[P.E = P.{E_1} + K.{E_1}\] …… (3)
Substituting the required values in the equation (3), we get:
$\Rightarrow P.E = P.{E_1} + K.{E_1} \\
\Rightarrow 1960 = 980 + K.{E_1} \\
\Rightarrow K.{E_1} = 1960 - 980 \\
\Rightarrow K.{E_1} = 980 \, {\text{J}} $

Hence, the kinetic energy of the object at half-way down is \[980 \, {\text{J}}\] .

Note: At the maximum height, the potential energy is always maximum. As the object falls, potential energy tends to decrease and kinetic tends to increase as the object gains more and more velocity due to acceleration due to gravity. When the object hits the ground, the kinetic energy is converted into sound energy and a part of it is also used to overcome the friction.