Question

write the statement of work energy theorem and prove it​

Answer 1

$<b>Work Energy Theorem$

According to the work energy theorem, the work done by a force on a moving body in the same direction is equal to the increase in its kinetic energy.

$<b>Proof$

Let a body of mass m be moving with an initial velocity u. When a constant force F is applied on the body along its direction of motion, it produces an acceleration a and the velocity changes from u to v in moving a distance S. Then

Force=mass×acceleration
F=ma .............. (1)

Work done by a force=Force×Displacement
W=F×S............. (2)

From relation v²=u²+2aS

$s = \frac{ {v}^{2} - {u}^{2} }{2a} .......(3)$

Substituting the value of F and S from equation (1) and (3) in equation (2), we get

$w = ma( \frac{ {v}^{2} - {u}^{2} }{2a}) \\ w = \frac{1}{2} m( {v}^{2} - {u}^{2} ) \\ w = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} ........(4)$

But initial kinetic energy
$k(initial) = \frac{1}{2} m {u}^{2}$

Final kinetic energy
$k(final) = \frac{1}{2} m {v}^{2}$

Then from equation 4

$w = k(final) - k(initial)$

Thus, work done on body=Increase in kinetic energy

Hence work energy theorem is proved.

Hope it helps.

Answer 2

$\bf \large {Work \: Energy \: Theorem}$

$\bf \large \color {blue}{Kinetic \: Energy \: and \: Work-Energy}$
$\bf \large \color {blue}{Theorem}$

The work-energy theorem states that the work done by all forces acting on a particle equals the change in the particle’s kinetic energy.

$\bf \color {blue}{Key\: takeaway}$

$\bf \color {purple}{Key \: points:-}$

1. The work W done by the net force on a particle equals the change in the particle’s kinetic energy KE:

W=ΔKE= $\frac{1}{2} {mv}^{2} \frac{}{f} - \frac{1}{2} {mv}^{2} \frac{}{i}$

2. The work-energy theorem can be derived from Newton’s second law.

3. Work transfers energy from one place to another or one form to another. In more general systems than the particle system mentioned here, work can change the potential energy of a mechanical device, the heat energy in a thermal system, or the electrical energy in an electrical device.

$\bf \large \color {purple}{Key \: Terms}$

$\bf {Torque:}$ A rotational or twisting effect of a force; (SI unit newton-meter or Nm; imperial unit foot-pound or ft-lb)

$\bf \large \color {blue}{The \: Work-Energy \: Theorem}$

The principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. This definition can be extended to rigid bodies by defining the work of the torque and rotational kinetic energy.

$\boxed {image\: is\: in \: the \: attachment\: above...}$

$\bf \large \color {blue}{Kinetic \: Energy:}$ A force does work on the block. The kinetic energy of the block increases as a result by the amount of work. This relationship is generalized in the work-energy theorem.

The work W done by the net force on a particle equals the change in the particle’s kinetic energy KE:

W=ΔKE= $\frac{1}{2} {mv}^{2} \frac{}{f} - \frac{1}{2} {mv}^{2} \frac{}{i}$

where $v\frac {}{i}$ and $v\frac {}{f}$ are the speeds of the particle before and after the application of force, and m is the particle’s mass.

$\bf \large \color {blue}{Derivation}$

For the sake of simplicity, we will consider the case in which the resultant force F is constant in both magnitude and direction and is parallel to the velocity of the particle. The particle is moving with constant acceleration a along a straight line. The relationship between the net force and the acceleration is given by the equation F = ma (Newton’s second law), and the particle’s displacement d, can be determined from the equation:

${v}^{2} \frac{}{f} = {v}^{2} \frac{}{i} + 2ad$

obtaining,

$d = \frac{v^{2} \frac{}{f} - {v}^{2} \frac{}{i} }{2a}$

The work of the net force is calculated as the product of its magnitude (F=ma) and the particle’s displacement. Substituting the above equations yields:

$w = fd = ma \frac{ {v}^{2} \frac{}{f} - {v}^{2} \frac{}{i} }{2a} = \frac{1}{2} mv^{2} \frac{}{f} = ke \frac{}{i} - ke \frac{}{i} =$ΔKE