The energy of a body is defined as the capacity to do work. It is a scalar quantity. Mechanical energy is one of the energy. There are two types of mechanical energy i.e. kinetic energy and potential energy.
The energy possessed by the body due to its motion is called kinetic energy. Few examples of bodies possessing kinetic energy are a moving bus, moving air, moving water in a river, a moving bullet etc.
Expression for Kinetic Energy
Let us consider a block of mass m lying on a smooth horizontal surface as shown in the figure. Let a constant force F be applied on it such that, after travelling a distance s, its velocity becomes v.
if a is the acceleration of the body, then
\begin{align*} v^2 &= u^2 + 2as = 0 + 2as \\ \text {or,} v^2 &= 2as \\ \text {initial velocity,} u=0. \text {So} \\ as &= \frac {v^2}{2} \dots {i} \\ \text {As work done on the body} &= \text {force} \times \text {distance, \: then}\\ W &= F\times s \\ &= mas \dots (ii) \\ \text {From equations} (i) \text {and} (ii), \text {we have} \\ W &= m \times \frac {v^2}{2} = \frac 12 mv^2 \\ \end{align*}
As this work done on the body of mass m will set the body moving with velocity v from its state of rest, it is equal to the kinetic energy of the body. Therefore,
$$\text {kinetic energy, K.E.} = \frac 12 mv^2 $$
It is the energy possessed by a body due to its position or due to its configuration. For example, as object above the ground, a compressed gas, a stretched rubber band, a compressed spring etc.
Gravitational Potential Energy
Consider a body of mass m on the surface of the earth. The earth attracts it with a force equal to its weight mg. if it is taken to a height h above the ground s shown in the figure, work done against the gravity is
$$ W = mg \times h $$
This work done will be stored in the body as gravitational potential energy.
When a spring or an elastic body is stretched, the certain amount of work done is stored in the spring in the form of elastic potential energy. Thus P.E. of a spring is the energy associated with the stated of compression or extension of an elastic spring.
To calculate it, consider an elastic spring OA of negligible mass. The end O of the spring is fixed to a rigid support and a body of mass m is attached to the free end A. let the spring be oriented along the axis and the body of mass m lie on a perfectly frictionless horizontal table.
The position of body A, when spring is compressed or elongated, it tends to recover its original length, on account of elasticity. The force trying to bring the spring back to its original configuration is called restoring force or spring force.
For a small stretch or compression, spring obeys Hooke’s law i.e. for a spring. Restoring force α stretch or compression
\begin{align*} \text {or,} \: -F &\propto x \\ \text {or,} \: -F &= kx \dots (i) \\ \end{align*}
where K is spring constant or force constant. The negative sign of equation (i) shows that the restoring force is directed always towards the equilibrium position. Let the body be displaced further through a small distance dx against the restoring force, small amount of work is done.
dw = -Fdx =kx dx
To obtain the total amount of work done, we have to integrate above equation from 0 to x.
\begin{align*} \text {i.e.} w &= \int _0^x kx dx \\ &= k\left [\frac {x^2}{2} \right ]_0^x = k\left [\frac {x^2}{2} – 0\right ] \\ \text {or,}\: w &= \frac 12 kx^2 \\ \text {This work done is stored in the spring s elastic energy or P.E. of the spring at point B.} \\ \therefore \text {P.E. at point B} &= \frac 12 kx^2 \\ \text {The variation of P.E. with distance x is shown in the figure} \\ \end{align*}
DISCUSSIONS ABOUT THIS NOTE
No discussion on this note yet. Be first to comment on this note