Prove work-energy theorem for constant work.

Hint: Work-energy theorem is work on the statement ‘change in energy of the body is equal to work done by the body’. The dimensional of the energy and the work is the same, which also shows that both the quantities are dimensionally equal. So the work and the kinetic energy can be written equal to each other.
Formula Used:
\[F=ma\]
\[\Delta W=F\centerdot d\]
\[{{v}^{2}}-{{u}^{2}}=2ad\]
\[K=\dfrac{1}{2}m{{v}^{2}}\].

Complete step-by-step solution:
Work-energy theorem: Work-done by a body or work done on a body is equal to change in its kinetic energy.
\[\Delta K=\Delta W\]
\[\Rightarrow {{K}_{i}}-{{K}_{f}}=\Delta W\]
Assume if the force applied on a body that has mass and due to this force it displaced distance. The initial velocity of the body is $u$ and the final velocity of the body is $v$. The motion of the body is as shown in the figure.


Here the constant force \[F\] is applied to the body of mass m. The body displaced by distance \[d\].
From the figure,
The well known Newton’s law of motion,
\[F=ma\]
The work done on the body is,
\[\Delta W=F\centerdot d\]
\[\Rightarrow \Delta W=Fd\cos \theta \]
\[\Rightarrow \Delta W=ma\times d\] (\[\because \cos 0=1\])
\[\Rightarrow \Delta W=ma\times \left( \dfrac{{{v}^{2}}-{{u}^{2}}}{2a} \right)\] (From the: \[{{v}^{2}}-{{u}^{2}}=2ad\] )
\[\Rightarrow \Delta W=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}\]
\[\Rightarrow \Delta W={{K}_{f}}-{{K}_{i}}\]
Hence proved that \[\Delta W={{K}_{f}}-{{K}_{i}}\].

Note:
(1) Kinetic energy is a scalar quantity. Hence is independent of its path and only depends on the body’s initial velocity magnitude and final velocity magnitude.
(2) The SI unit of work and energy is the same, and it is \[Nm\] also often denoted by\[Joule(J)\].
(3) The work-energy theorem is the part of the conservation of energy where the potential energy of the body is absent and no external forces (for example friction) present in the motion.