we already have an idea that energy is associated closely with work and we have defined energy of a body as the capacity of the body to do work
In dynamics body can do work either due to its motion ,due to its position or both due to its motion and position
Ability of a body to do work due to its motion is called Kinetic energy for example piston of a locomotive is capable of doing of work
Ability of a body to do work due to its position or shape is called Potential Energy
For example workdone by a body due to gravity above surface of earth
Sum of kinetic energy and Potential energy of body is known as its mechanical energy
Thus
Mechanical Energy=Kinetic Energy+Potential Energy
(4) Principal of Conservation of Mechanical Energy
From work energy theorem , we know that
$\Delta K= W_{net}$
Now for conservative forces , we know that
$\Delta = - F.dx$
or
$W = -\Delta U$
If only conservative forces acting on the system ,then
$\Delta K= W_{net}$
$\Delta K = - \Delta U$
$ \Delta K + \Delta U=0$
or
$K_2 - K_1 + U_2 - U_1=0$
$K_2 + U_2 = K_1 + U_1$
or
$K + U=constant$
We already know that above quantity is called the mechanical energy of the system
So we see that if only conservative forces are acting on the system, the total mechanical energy of the system remains constant.It does not increase or decrease and it is conserved . This is called the Principal of Conservation of Mechanical Energy
If non-conservative forces are also present in the system such as friction, the Work Energy Theorem is given as
$W_{net}= W_{c} + W_{nc}$
Now ,$\Delta K= W_{net}$
Therefore,
$\Delta K=W_{c} + W_{nc}$
$\Delta K -W_{c}= W_{NC}$
$\Delta K + \Delta U = W_{NC}$
So, we see that total mechanical energy is not conserved if the non-conservative forces are present
Example-1
A man throws an ball of mass .10 kg from the top of the building of height 10 m with speed of 20 m/s. Find kinetic energy and speed of the ball when it reaches the ground? (take g=10 m/s^{2}) Solution
From law of conservation of mechanical energy
(Kinetic Energy + Potential energy) at top = ( Kinetic energy ) at ground
$KE_{ground}= \frac {1}{2} mv^2 + mgH = \frac {1}{2} .1 20^2 + .1 \times 10 \times 10= 30 J$
Now
$\frac {1}{2} mv_g^2 = 30$
or
v=24.5 m/s
Example-2
A object of mass 10 kg moving with a speed 5 m/s on a smooth surface and it collide with a horizontally mounted spring of spring constant 1000 N/m. What is the
maximum compression of the spring ? Solution
At maximum compression , Kinetic energy of the object converts into potential energy of the spring
From law of conservation of mechanical energy
(Kinetic Energy) initially = ( Potential Energy ) at maximum compression
$\frac {1}{2}mv^2 = \frac {1}{2}kx^2$
or
$ x= v \sqrt {\frac {m}{k}}$
or x=.5 m