Derive an expression for kinetic energy of a mass m, moving with velocity v.

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Accepted Answer

>

> [!TLDR]
> Using the work-energy theorem and Newton's second law, the work done to bring a mass m from rest to speed v works out to exactly (1/2)mv^2 — defined as its kinetic energy.
>
> **Step 1: Set up the problem.**
> Consider a body of mass m, initially at rest, acted on by a constant force F along the direction of motion, which accelerates it uniformly to a final speed v after travelling distance s. We want to find the work done by F, since by the work-energy theorem this equals the kinetic energy gained.
>
> **Step 2: Start from Newton's second law.**
> F = ma
>
> **Step 3: Write the work done as an integral.**
> W = ∫F ds (from initial position 0 to final position s)
> W = ∫ma ds
>
> **Step 4: Use the identity a = v(dv/ds).**
> Since a = dv/dt = (dv/ds)(ds/dt) = v (dv/ds), substitute:
> W = ∫m v (dv/ds) ds = ∫m v dv
>
> **Step 5: Integrate from the initial speed (0) to the final speed (v).**
> W = m ∫₀^v v dv = m [v^2/2]₀^v = (1/2)mv^2
>
> **Step 6: Apply the work-energy theorem.** …

Speed change	Kinetic energy change
Double speed ( $2v$ )	$K$ becomes $4 \times$ original
Triple speed ( $3v$ )	$K$ becomes $9 \times$ original
Halve speed ( $v/2$ )	$K$ becomes $1/4$ of original

Q.Derive an expression for kinetic energy of a mass m, moving with velocity v.

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Q.Derive an expression for kinetic energy of a mass m, moving with velocity v.

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