Question

Name the unit of physical quantity obtained by the formula $\dfrac{2K}{{{V}^{2}}}$. Where, K: kinetic energy, V: linear velocity.

Answer 1

Hint: We are given a formula in terms of kinetic energy and linear velocity. As a first step, you could recall the expression for kinetic energy. Then, you could substitute this in the given formula. After that you could cancel out the common terms on the numerator and denominator and finally you will end up with a physical quantity, which would be the answer.

Formula used:
Kinetic energy,
$K=\dfrac{1}{2}m{{V}^{2}}$

Complete Step by step solution:
In the question, we are asked to find the physical quantity that is obtained from the formula given by $\dfrac{2K}{{{V}^{2}}}$
We are also told in the question that K represents kinetic energy and V represents the linear velocity.

You may recall that the expression for kinetic energy is given by,
$K=\dfrac{1}{2}m{{V}^{2}}$
Where, m is the mass and V is the linear velocity.

Now we could substitute this expression in the given formula in place of the kinetic energy K to get,
$\dfrac{2K}{{{V}^{2}}}=\dfrac{2\times \dfrac{1}{2}m{{V}^{2}}}{{{V}^{2}}}$
$\therefore \dfrac{2K}{{{V}^{2}}}=m$

Therefore, we found that the physical quantity represented by the given formula in the question is mass (m).

Additional information:
Mass can be defined in two ways. It could be defined as the property of the physical body in one way or as the measure of resistance offered to the acceleration of the body under the action of a net force. It is the object’s mass that determines how strongly it is attracted by other bodies under the influence of gravitational force. The SI unit of mass is known to be kilogram (kg).

Note:
Though a common man may consider mass to be the same as the weight of the body, as a physicist, you should know that, they both are two entirely different concepts. Mass and weight are related to each other by the following relation,
$W=mg$
As acceleration due to gravity is different on earth and moon, the weight of the object would be different on both but the mass will remain constant on both.