Question

dimensions of kinetic energy​

Answer 1

Kinetic energy is the amount of energy that an item or particle has as a result of its motion. When work is done on an item by transferring energy via a net force, the object accelerates and absorbs kinetic energy. Kinetic energy is a characteristic of a moving item or particle that is determined by its mass and velocity. The kind of motion might be translation (or movement along a path from one point to another), rotation around an axis, acceleration, or any combination of movements.

Derivation

Kinetic energy (K.E) = [Mass × Velocity2] × 2-1 ---------------- (1)

The dimensional formula of Mass = [M1 L0 T0] --------------(2)

Subsequently, Velocity = Distance × Time-1 = L × T-1

Therefore, The dimensional formula of velocity = [M0 L1 T-1] --------------------(3)

Now substitute equation (2) and (3) in equation (1) then we get,

⇒ K.E= [Mass × Velocity2] × 2-1

Or, K.E = [M1 L0 T0] × [M0 L1 T-1]2 = [M1 L2 T-2]

If kinetic energy is the energy of motion then the kinetic energy of an object in rest should naturally be zero. So we don't need the second term and the kinetic energy of an object is just

Hence, Kinetic Energy is dimensionally expressed as [M1 L2 T-2].

Answer 2

We know that,

The formula for kinetic energy is;

$Kinetic energy=\frac{1}{2} Mv^{2}$

$=>\text { Kinetic energy }(K . E)=\left[\text { Mass } \times \text { Velocity }^{2}\right] \times 2^{-1} \ldots \ldots(1)\\\text =>{ The dimensional formula of Mass }=\left[\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \ldots(2)\\\text =>{ Since, Velocity }=\text { Distance } \times \text { Time }^{-1}=[L] \times[T]^{-1}\\\therefore \text { The dimensional formula of velocity }=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right] \ldots(3)$

$\text { On substituting equation (2) and (3) in equation (1) we get, }\\\\\Rightarrow \text { Kinetic energy }=\left[\text { Mass } \times \text { Velocity }^{2}\right] \times 2^{-1}$

$\text { Or, K.E }=\left[M^{1} L^{0} T^{0}\right] \times\left[M^{0} L^{1} T^{-1}\right]^{2}=\left[M^{1} L^{2} T^{-2}\right]\\\\\text { Therefore, Kinetic Energy is }\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right] .$

$\text { Hence, Kinetic Energy is represented as }\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right] .$

Where,

M = Mass

L = Length

T = Time