## Work energy theorem

Consider a constant force $\vec{F}$ acting on a particle of mass $\mathrm{m}$. Let a acceleration ‘a’ be produced in the direction of force $\overrightarrow{\mathrm{F}},$ say along $\mathrm{X}$ -axis.Let the resultant force vary in magnitude only, not in direction. The work done by the resultant force in displacing the particle from $x_{0}$ to $x$ is

$W=\int_{i}^{f} \vec{F} \cdot \overrightarrow{d x}=\int_{i}^{f} F \cdot d x$

Since F = ma, and

$\mathrm{a}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

= $\frac{d v}{d x} \cdot \frac{d x}{d t}$

$=v \frac{d v}{d x}$

Hence,

$W=\int_{i}^{f} m a d x$

$=\int_{i}^{f} m v \frac{d v}{d x} \cdot d x$

$=m \int_{i}^{f} v \cdot d v=m\left(\frac{v^{2}}{2}\right)_{i}^{i}$

Or

$W=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}$

**“The work done by the resultant force acting on a particle is equal to the change in the kinetic energy of the particle”.**

$W=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}$

**This is Work Energy Theorem.**

Let a particle of mass m is moving with velocity v. This particle is stopped by some resistive force i.e. stopping force and due to action of this force particle comes in rest.

Then

$W=0-\frac{1}{2} m v^{2}$

$=-\frac{1}{2} \mathrm{mv}^{2}$

Negative sign shows that particle is doing work against the stopping force. It means capacity of doing

work of a moving particle is $\frac{1}{2} \mathrm{mv}^{2}$. So that it is known as kinetic energy of particle.

$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$

The energy possessed by a body by virture of its motion is known as kinetic energy.

The work done on a particle by the resultant force is equal to the change in Kinetic energy of the particle.

So, the work done by the resultant force

$W \quad=K_{f}-K_{i}=\Delta K$

**Salient features of the work-energy theorem**

**Kinetic energy**

The KE of a moving body is equal to the amount of work that must be done to bring a body from rest into the state of motion. Conversely, the amount of work that we must do in order to bring a moving body to rest is equal to the negative of the kinetic energy of the body, i.e.,
KE = work done to put the body into motion

= –work done to bring the body to a stop

As mass m and $\mathrm{v}^{2}(\overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{v}})$ are always positive, kinetic energy is always positive scalar, i.e., kinetic energy

can never be negative.

KE is always positive. The KE is a scalar quantity.

- Suppose a block is at rest on a frictionless surface and a constant net force F acts on the block. If v be velocity acquired by the block after travelling a distance x, then KE is $\mathrm{K}=\mathrm{W}=\mathrm{Fx}=\mathrm{m} \cdot \mathrm{a} \cdot \mathrm{x}=\frac{1}{2} \mathrm{mv}^{2}$ $\left[\because v^{2}=2 \mathrm{ax}\right]$
- If the block is already in motion when the constant force is applied to it, then the work done is equal to change in KE of the block, i.e.

$\Delta K=K_{4}-K_{1}=W=F x$

$m \cdot a \cdot x=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$ - KE depends on the frame of reference, e.g., KE of a person of mass $\mathrm{m}$ sitting in a train moving with speed $v$ is zero w.r.t. frame of train but $(1 / 2) \mathrm{mv}^{2}$ w.r.t. frame of reference of earth.

So, that’s all from this Topic. I hope you liked the Explanation of Work energy theorem Derivation. If you have any confusion related to this topic please let us know in the comments section down below.

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