Class 9, Science, Chapter-10, Lecture-2, Energy (Notes)

Kinetic Energy:

The energy possessed by a body by virtue of its motion is termed as Kinetic Energy. 

Mathematically:
${{\rm{E}}_{\rm{K}}} = {1 \over 2}{\rm{m}}{{\rm{v}}^{\rm{2}}}$

where,
${{\rm{E}}_{\rm{K}}} = {\rm{ ~Kinetic~ Energy~ of~ a ~moving ~object}}$
${\rm{m}} = {\rm{~Mass ~of~ the~ moving~ object}}$
${\rm{v}} = {\rm{ ~Velocity ~of~ the~ moving~object}}$

Derivation of ${{\rm{E}}_{\rm{K}}} = {1 \over 2}{\rm{m}}{{\rm{v}}^{\rm{2}}}$:

Let ‘F’ be the continuous constant force acting on an object of mass ‘m’ which is initially at rest. 
Let ‘s’ be the displacement caused by the force.
Let the acceleration caused by the force be ‘a’ and the final velocity be ‘v’. 

The work done on the body by the force ‘F’ is 
${\rm{W}} = {\rm{F}} \cdot {\rm{s}}$ $ - \,\,\, - \,\,\, - \,\,\,(i)$
From Newton’s Second Law 

${\rm{F}} = {\rm{m}} \cdot {\rm{a}}$

From Third Equation of Motion 

${{\rm{v}}^{\rm{2}}} - {{\rm{u}}^{\rm{2}}} = 2{\rm{as}}$
$ \Rightarrow $${{\rm{v}}^{\rm{2}}} = 2{\rm{as}}$
$ \Rightarrow $${\rm{s}} = {{{{\rm{v}}^{\rm{2}}}} \over {2{\rm{a}}}}$$ - \,\,\, - \,\,\, - \,\,\,(iii)$

Using (ii) and (iii) in (i) 

${\rm{W}} = \left( {{\rm{m}} \cdot {\rm{a}}} \right) \cdot \left( {{{{{\rm{v}}^{\rm{2}}}} \over {{\rm{2a}}}}} \right)$
${\rm{W}} = {1 \over 2}{\rm{m}} \cdot {{\rm{v}}^{\rm{2}}}$
Since this workdone is the change in kinetic energy of the body. Therefore
${\rm{KE}} = {1 \over 2}{\rm{m}}{{\rm{v}}^{\rm{2}}}$

Potential Energy: 

The energy possessed by a body by virtue of its position or shape is termed as Potential Energy. 

Mathematically:

${{\rm{E}}_{\rm{P}}} = \,\,{\rm{m}}\,{\rm{g}}\,{\rm{h}}$

where,

${{\rm{E}}_{\rm{P}}} = {\rm{ ~Potential~ Energy~ of~ a ~moving~ object}}$ 
${\rm{m}} = {\rm{~ Mass ~of~ the ~moving~ object}}$
${\rm{g}} = {\rm{ Acceleration ~due~ to ~gravity}}$
${\rm{h}} = {\rm{ ~Height ~of ~object~ from ~the ~surface~ of~ the ~earth}}$

Derivation of ${{\rm{E}}_{\rm{P}}} = \,\,{\rm{m}}\,{\rm{g}}\,{\rm{h}}$:

Let ‘m’ be the mass of an object placed at the surface of the earth. 
Let ‘h’ be the height to which the object is lifted.
Let ‘g’ be the acceleratation due to gravity.
The force required to lift the object without producing acceleration is equal to the weight of the object.
$\therefore{\rm{F}} = {\rm{m}}\,{\rm{g}}$
The work done on the body by the force ‘F’ is 
${\rm{W}} = {\rm{F}} \cdot {\rm{h}}$ ${{\rm{h~ is ~the ~displacement~ of~ the ~object~ in ~the~ direction~ of ~applied ~force}}}$
${\rm{W}} = {\rm{m}}\,\,{\rm{g}}\,\,{\rm{h}}$
Since this workdone is the change in potential energy of the body. Therefore
${{\rm{E}}_{\rm{P}}} = \,\,{\rm{m}}\,{\rm{g}}\,{\rm{h}}$