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Physical Science:

Dr. Sunil Karna

Subsection 3.1.1 Work-Energy Theorem

Let us apply Newton’s II law into II equation of motion and rearange them.
\begin{align*} v^2_f \amp = v^2_i + 2 a d\\ or, v^2_f \amp = v^2_i + 2 \frac{F}{m} d\\ or, v^2_f - v^2_i \amp = 2 \frac{Fd}{m} \\ or, m\left(v^2_f - v^2_i\right) \amp = 2 F d\\ or, \frac{1}{2} \left(mv^2_f - mv^2_i\right)\amp = Fd\\ or, \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\amp = W \end{align*}
\begin{equation} \therefore W = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\tag{3.1.5} \end{equation}
Here, \(\frac{1}{2}mv^2_f \) is final kinetic energy and \(\frac{1}{2}mv^2_i \) is initial kinetic energy. Hence, \(\frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\) is change in kinetic energy. Therefore, work done = change in kinetic energy. or
\begin{equation*} W = \Delta KE \end{equation*}
This is known as work-energy theorem. Meaning the amount of work is done on a body is equal to its change in kinetic energy or vice a versa along the straight line.

Example 3.1.3.

A block of mass \(2 kg\) at rest is being pushed by an unknow force along the horizontal to bring it at 5 \(m/s\) velocity.
  1. How much work is done by the force?
  2. If the body moves to a 2 m distance in this process. Find the value of unkown force.
Solution.
Given: \(m=2 kg\text{,}\) \(v_i = 0\text{,}\) \(v_f = 5 m/s\text{,}\) \(d=2m\text{.}\)
  1. \(W=?\)
    \begin{align*} W \amp = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\\ or, W \amp = \frac{1}{2}2\cdot5^2-0\\ \therefore W \amp = 25 J. \end{align*}
  2. \(F=?\)
    \begin{align*} W \amp = Fd\\ or, F \amp = \frac{W}{d}\\ \therefore F \amp = \frac{25}{2} = 12.5 N. \end{align*}
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