Work Energy and Power
5.0 Kinetic Energy $(K)$
5.0 Kinetic Energy $(K)$
Kinetic energy is the capacity of a body to do work by virtue of its motion.
Kinetic energy for a body of mass $m$ moving with velocity $v$ is given by $\frac{1}{2} mv^2$
For better understanding, let us derive the above equation.
Suppose a block of mass $m$ is subjected to an external force $F$ in the horizontal direction. The block is free to slide on a horizontal frictionless surface.
Work done by all the force is equal to the gain in kinetic energy by the block. So, $$\begin{equation} \begin{aligned} W = \int {\overrightarrow F } .d\overrightarrow s \\ W = \int {ma(ds)} \\ W = \int {m\frac{{dv}}{{dt}}ds} \\ W = \int {m\frac{{ds}}{{dt}}dv} \\ W = \int {mvdv} \\ W = \frac{1}{2}m{v^2} \\\end{aligned} \end{equation} $$ or $$K = \frac{1}{2}m{v^2}$$
Note:
1. Kinetic energy is a scalar quantity and is always positive. It does not depend on the direction of motion of the body. $$K = \frac{1}{2}m\overrightarrow v .\overrightarrow v = \frac{1}{2}m{\left| v \right|^2}$$
${\left| v \right|^2}$: This term is independent of the direction as it is square of the magnitude.
We also know that kinetic energy is a dot product of two vector quantities. So, it is a scalar quantity.
2. Kinetic energy $(K)$ depends on the frame of reference.
3. Relation between kinetic energy $(K)$ and linear momentum $(p)$.
Linear momentum of mass $m$ moving with velocity $v$ is given as, $$\overrightarrow p = m \overrightarrow v ...(i)$$
and $$K = \frac{1}{2}m{v^2}...(ii)$$
From equation $(i)$ & $(ii)$ we get,$$K = \frac{{{p^2}}}{{2m}}$$ or $$p = \sqrt {2mK} $$
4. Graph
 ? |   ? |  ? |
| For $K = \frac{{{p^2}}}{{2m}}$, graph between $K$ and $p$ is a parabola. | For $p = \sqrt {2mK} $, graph between $\sqrt{K}$ and $p$ is a straight line. | For $\sqrt K \left( {\frac{1}{p}} \right) = \frac{1}{{\sqrt {2m} }}$, graph between $\sqrt{K}$ and $\frac{1}{p}$ is a hyperbola. |
Question 11. A ball of mass $m$ is thrown vertically upwards with an initial velocity $u$. Find the kinetic energy ,
- at the initial point $(t=0)$
- after time $t$
- when the ball is at the highest point
Solution:
1. A the initial point $(t=0)$
Initial velocity $ = u\widehat j$
Kinetic energy, ${KE_1}=\frac{1}{2}m{u^2}$
2. After time $t$
Initial velocity $ = u\widehat j$
Acceleration $ = -g\widehat j$
From the equation $\overrightarrow v = \overrightarrow u + \overrightarrow a t$, we get, $$\overrightarrow v = (u-gt)\widehat j$$
Kinetic energy $(K{E_2})$, $$\begin{equation} \begin{aligned} K{E_2} = \frac{1}{2}m{v^2} = \frac{1}{2}m{\left( {u - gt} \right)^2} \\ \therefore K{E_2} = \frac{1}{2}m{\left( {u - gt} \right)^2} \\\end{aligned} \end{equation} $$
3. When the ball is at the highest point
At highest point the velocity $\overrightarrow v$ of the ball is zero. So, kinetic energy is also zero. $$\therefore K{E_3} = \frac{1}{2}m{\left( 0 \right)^2} = 0$$
