Question

A body of mass 200 g begins to fall from a height where its potential energy is 80J. Its velocity at a point where kinetic and potential energies are equal is:
\eqalign{ & {\text{A}}{\text{. }}4m/s \cr & {\text{B}}{\text{. }}400m/s \cr & {\text{C}}{\text{. }}20m/s \cr & {\text{D}}{\text{. }}10\sqrt 8 m/s \cr}

Answer 1

When a body falls down from any height on earth, the force of gravity acts on it. Find the expression of total energy in terms of potential energy. Later, substitute the condition of kinetic energy and potential energy are equal in this expression to obtain the value of velocity.

Formula Used:
The total energy of the system, $T.E.{\text{ }} = P.E. + K.E.$
Kinetic Energy of body, $K.E = \dfrac{1}{2}m{v^2}$

Complete step-by-step answer:
The potential energy, P.E. of a body is defined as the energy held inside an object due to its position relative to other objects, its electric charge, stresses within itself, or other similar factors. An object tends to possess gravitational potential energy if it is positioned at a height above the ground. For a body of mass, m at a certain height, h above the ground the potential energy, P.E. is given mathematically as:
$P.E. = mgh{\text{ }} \cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Kinetic energy is the energy possessed by a moving object due to its motion. In order words, it can be defined as the work needed to accelerate a body of given mass from rest to its stated velocity.
For a body of mass, m moving with velocity, v the kinetic energy, K.E. is given mathematically as:
$K.E = \dfrac{1}{2}m{v^2}{\text{ }} \cdots \cdots \cdots \cdots \cdots \left( 2 \right)$
The total energy of an object is the sum of its potential energy and kinetic energy, given that no external force is acting on it.
\eqalign{ & {\text{Total energy = Potential energy + Kinetic energy}} \cr & T.E.{\text{ }} = P.E. + K.E. \cr}
$T.E. = mgh + \dfrac{1}{2}m{v^2}$
Given:
The mass of the body, $m = 200g = 0.2kg$
The potential energy of the body, $P.E = 80J$
Now, because the body falls under gravitational force which is conservative in nature, so the total energy of the body initially must be equal to the total energy throughout the motion by law of conservation of energy.
\eqalign{ & \Rightarrow T.E. = P.E\left( {{\text{initially}}} \right) \cr & \Rightarrow T.E. = 80J{\text{ }}\left[ {\because P.E. = 80J} \right] \cr}
Now, we wish to find the velocity of the body at a point where potential energy is equal to the kinetic energy of the object.
\eqalign{ & i.e.{\text{ }}P.E. = K.E \cr & \because T.E. = P.E. + K.E \cr & \Rightarrow 80J = 2K.E. \cr}
From equation (2), we get:
\eqalign{ & 80J = 2\left( {\dfrac{1}{2}m{v^2}} \right) \cr & \Rightarrow 80J = m{v^2} \cr & \Rightarrow {v^2} = \dfrac{{80}}{{0.2}} \cr & \Rightarrow {v^2} = 400 \cr & \therefore v = \sqrt {400} = 20m{s^{ - 1}} \cr}
Therefore, the correct option is C. i.e., the velocity of the body at a point where kinetic and potential energies are equal is $20m/s$

Note: Students can make calculation errors by not converting the mass of the body which is given in grams into kilograms. It is very crucial to follow only one unit system to avoid incorrect answers.Additionally, the total energy of the system remains conserved, i.e., the initial energy and the final energy of the system remain the same because it acts under the force of gravitation which is a conservative force.