Question

A body of mass \[2\,{\text{kg}}\] moves down the quadrant of a circle of radius
\[4\,{\text{m}}\] . The velocity on reaching the lowest point is \[8\,{\text{m}}{{\text{s}}^{ - 1}}\] .Then the loss of energy is:
A. \[78.4\,{\text{J}}\]
B. \[64\,{\text{J}}\]
C. \[14.4\,{\text{J}}\]
D. Zero

Answer 1

Hint: First of all, we will calculate the potential energy of the body which is possessed by the body by virtue of the position and then we will calculate the kinetic energy of the body. Then we will find the difference.

Complete step by step answer:
Given,
The mass of the body is \[2\,{\text{kg}}\] .
The radius of the circle is \[4\,{\text{m}}\] .
The velocity at the lowest point is \[8\,{\text{m}}{{\text{s}}^{ - 1}}\] .
We are asked to find the loss of energy.
To begin with, we must understand the situation that at the beginning the body is located at the point on the top of the circle, whose radius is provided to us. Then the body moves down whose velocity at the lowest point is also provided. At the highest point the body has potential energy in it by the virtue of its mass and height. Then when the body moves to a lower point, the body has kinetic energy in it due to its motion. The energy lost is due to overcoming the friction between the surface and the body.

So first, we will calculate the potential energy of the body, which is given by the equation, as given below:
\[P.E = mgh\]
Since, the position of the body can also be said as the radius of the circle. So, we can modify the equation as:
\[P.E = mgr\] …… (1)
Where,
\[P.E\] indicates the potential energy.
\[m\] indicates the mass of the body.
\[g\] indicates the acceleration due to gravity.
\[r\] indicates the radius of the circle.
Now, we substitute the required values in the equation (1), and we get:
$
P.E = mgr \\
\Rightarrow P.E = 2 \times 9.8 \times 4 \\
\Rightarrow P.E = 78.4\,{\text{J}} \\
$
The potential energy is found to be \[78.4\,{\text{J}}\] .

Now, we will find the kinetic energy, which is given by the following expression:
\[K.E = \dfrac{1}{2}m{v^2}\] …… (2)
Where,
\[K.E\] indicates the potential energy.
\[m\] indicates the mass of the body.
\[v\] indicates the velocity of the body.
Substituting the required values in the equation (2), we get:
$
K.E = \dfrac{1}{2}m{v^2} \\
\Rightarrow K.E = \dfrac{1}{2} \times 2 \times {8^2} \\
\Rightarrow K.E = 64\,{\text{J}} \\
$
The kinetic energy is found to be \[64\,{\text{J}}\].

Now, the loss in energy can be calculated as:
$
\Delta E = P.E - K.E \\
\Rightarrow \Delta E = 78.4 - 64 \\
\Rightarrow \Delta E = 14.4\,{\text{J}} \\
$
Hence, the loss in energy of the body is found out to be \[14.4\,{\text{J}}\].

The correct option is C.

Note: It is important to note that a body if it is rest above a certain point from the ground, then it possesses potential energy by the virtue of its position. Many students, while solving this problem, just calculate the kinetic energy of the body, which is wrong. We have to calculate the difference, which is the actual loss in energy.