Work energy theorem Derivation - Easy Step by Step Explanation.

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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

• When the speed of the particle is constant, there is no change in Kinetic energy and the work done by the resultant force is zero. For example, in case of uniform circular motion, the speed of the particle is constant and so the centripetal force does no work on the particle.
• It can also be concluded that if no external work is done, the Kinetic energy before a process must be equal to its Kinetic energy at the conclusion of the process.
• Kinetic energy of a particle decreases by an amount just equal to the amount of work which the particle performs. A body is said to have energy associated with it because of its motion; as it does work it slows down and loses some of this energy.
• Work and energy are interchangeable quantities. When work is done, it appears as energy. The energy can be decreased by permitting the particle to do work on other particles.

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.

1. 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 $\mathrm{K}=\mathrm{W}=\mathrm{Fx}=\mathrm{m} \cdot \mathrm{a} \cdot \mathrm{x}=\frac{1}{2} \mathrm{mv}^{2}$ $\left[\because v^{2}=2 \mathrm{ax}\right]$
2. 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.
   
   $\Delta K=K_{4}-K_{1}=W=F x$
   
   $m \cdot a \cdot x=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$
3. 
   
4. 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.

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