Work Energy and Power
8.0 Power
8.0 Power
Power is defined as the rate of doing work. If $\Delta W$ is the amount of work performed during time $\Delta t$, the average power $P_{avg}$ over the time $\Delta t$ is given by the formula, $$P_{avg}=\frac{\Delta W}{\Delta t}$$
The "average power" is simply called as power.
Instantaneous power $(P_{ins})$
Power at any instant is defined as the instantaneous power.
Instantaneous power is the limiting value of the average power when the time $\Delta t$ approaches zero. $$\begin{equation} \begin{aligned} {P_{ins}} = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta W}}{{\Delta t}} \\ {P_{ins}} = \frac{{dW}}{{dt}} \\\end{aligned} \end{equation} $$
Instantaneous power is also defined as the dot product of the force $(F)$ and velocity $(v)$ at that instant. $$\begin{equation} \begin{aligned} {P_{ins}} = \overrightarrow F .\frac{{d\overrightarrow r }}{{dt}} \\ {P_{ins}} = \overrightarrow F .\overrightarrow v \\\end{aligned} \end{equation} $$
Note:
- SI unit of power is watt $(W)$, which is equal to one joule per second.
- CGS unit of power is $erg s^{-1}$ and $1W=10^7 erg s^{-1}$
- 1 Horsepower $(hp)$ is equal to $746 watts$
- Bigger units of power
- 1 killowatt = $1kW=10^3W$
- 1 megawatt=$1MW=10^6W$
- Power is a scalar quantity as it is a dot product of two vector quantities.
- Dimensional formula of power is $\left[ {M{L^2}{T^{ - 3}}} \right]$
Efficiency $\left( \eta \right)$
Efficiency is defined as the ratio of output energy to the input energy. It is represented by eta$\left( \eta \right)$. $$\eta = \frac{OutputEnergy}{InputEnergy}$$
For an ideal machine, $\eta=1$ which is equal to 100% efficiency.
Example:
An emulsion rod, which is, supplied $100J$ as input energy and it produces $40J$ of heat as output energy.
So, the efficiency of the emulsion rod is, $$\begin{equation} \begin{aligned} \eta = \frac{{OutputEnergy}}{{InputEnergt}} \\ \eta = \frac{{60}}{{100}} \\ \eta = 0.60 \\\end{aligned} \end{equation} $$
Mechanical advantage $(M.A.)$
Mechanical advantage is the ratio of the output force to th
e input force. $$M.A.=\frac{OutputForce}{InputForce}$$
Question 19. A machine does $400J$ of work over an interval of 25 $seconds$. Find the power delivered by the machine. Also, find the efficiency of the machine if $600J$ of electricity was supplied for the entire process.
Solution: We know, $$\begin{equation} \begin{aligned} W = 400J \\ t = 25\sec onds \\ P = \frac{W}{t} = \frac{{400}}{{25}} \\ P = 16watts \\\end{aligned} \end{equation} $$
Efficiency of the machine, $$\begin{equation} \begin{aligned} \eta = \frac{{OutputEnergy}}{{InputEnergy}} \\ \eta = \frac{{400}}{{600}} \\ \eta = 0.67 \\\end{aligned} \end{equation} $$
Question 20. A constant force $\overrightarrow F = \left( {2\widehat i + 3\widehat j + 4\widehat k} \right)N$ is applied on a block of mass $2kg$. The position of block varies with respect to time as $\overrightarrow r = \left( {3{t^2}\widehat i + 2t\widehat j + 3\widehat k} \right)m$. Find the power at $t=2s$.
Solution: $$\overrightarrow r = \left( {3{t^2}\widehat i + 2t\widehat j + 3\widehat k} \right)m$$
Differentiating with respect to $t$ we get, $$\begin{equation} \begin{aligned} \frac{{d\overrightarrow r }}{{dt}} = 6t\widehat i + 2\widehat j \\ \overrightarrow v = 6t\widehat i + 2\widehat j \\\end{aligned} \end{equation} $$
At, $t=2s$, $$\overrightarrow v = 12\widehat i + 2\widehat j$$
Therefore, $$\begin{equation} \begin{aligned} P = \overrightarrow F .\overrightarrow v \\ P = \left( {2\widehat i + 3\widehat j + 4\widehat k} \right).\left( {12\widehat i + 2\widehat j} \right) \\ P = \left( {24 + 6} \right) \\ P = 30watts \\\end{aligned} \end{equation} $$
