This is the product of the force applied to an object and the distance the object moves in the direction of the force: $$ \begin{equation}\begin{aligned} work&=force\times distance\\ &=Fd\\ \end{aligned}\end{equation} $$ The unit for work is the Joule(J): $$ \begin{equation}\begin{aligned} work=Fd\rightarrow\ N\times m\rightarrow Nm\ OR\ J\\ \end{aligned}\end{equation} $$ It has the same unit as energy since energy is the ability to do work. Note that the Newton-metre ($Nm$) is comprised of the fundamental units: $$ \begin{equation}\begin{aligned} Nm\rightarrow kgms^{-2}\times m\rightarrow kgm^2s^{-2}\equiv J\\ \end{aligned}\end{equation} $$
Example
A $5\ N$ force acting eastward is used to push a box by a distance of $7\ m$ in the east direction. Find the work done in order to move the box, assuming that the frictional force is negligible. $$ \begin{equation}\begin{aligned} work&=Fd\\ &=5\ N\times 7\ m\\ &=35\ Nm\ OR\ 35\ J\\ \end{aligned}\end{equation} $$
Power
This is the energy converted or transferred per unit time: $$ \begin{equation}\begin{aligned} power&=\frac{energy}{time}\\ P&=\frac{E}{t}\\ \end{aligned}\end{equation} $$ Looking at the units: $$ \begin{equation}\begin{aligned} P=\frac{E}{t}\rightarrow \frac{J}{s}\rightarrow Js^{-1}\ OR\ Watts(W)\\ \end{aligned}\end{equation} $$
The kilowatt-hour
This is a unit used to measure the amount of energy used. It is NOT a measure of power. $$ \begin{equation}\begin{aligned} P=\frac{E}{t}\rightarrow E&=Pt\\ \therefore E&=Pt\rightarrow kW\times h\rightarrow kWh\\ \end{aligned}\end{equation} $$ Thus the kilo-watt hour is the energy used if we run an appliance with a power rating of $1\ kW(1000\ W)$ for $1\ hr(3600\ s)$: $$ \begin{equation}\begin{aligned} 1\ kWh&\rightarrow 1\ kW\times 1\ h\\ &=1,000\ W\times 3,600\ s\\ &=3,600,000\ Ws\\ &=3.6\times 10^6\ J\\ &=3.6\ MJ\\ \end{aligned}\end{equation} $$
A kilowatt-hour is the same as 3.6 megajoules of energy.
Mechanical advantage (MA)
This is the ratio of the load to the effort. This quantity is dimensionless (no units): $$ \begin{equation}\begin{aligned} MA&=\frac{load}{effort}\color{gray}\rightarrow \frac{N}{N}\rightarrow no\ units\\ \end{aligned}\end{equation} $$
Example
If a lever requires a input force of $1\ N$ to move a box of weight $5\ N$: $$ \begin{equation}\begin{aligned} MA&=\frac{load}{effort}\\ &=\frac{5\ N}{1\ N}\\ &=5\\ \end{aligned}\end{equation} $$
Velocity Ratio (VR)
This is the ratio of the distance moved by the effort to the distance moved by the load: $$ \begin{equation}\begin{aligned} VR=\frac{distance\ moved\ by\ effort}{distance\ moved\ by\ load}\\ \end{aligned}\end{equation} $$
Example
In the case of the lever previously mentioned, the effort would have to move by $5\ m$ in order to move the load by only $1\ m$ (if the system is 100% efficient): $$ \begin{equation}\begin{aligned} VR&=\frac{distance\ moved\ by\ effort}{distance\ moved\ by\ load}\\ &=\frac{5\ m}{1\ m}\\ &=5\\ \end{aligned}\end{equation} $$
Efficiency
This is the percentage ratio of the output to the input: $$ \begin{equation}\begin{aligned} efficiency&=\frac{output}{input}\times 100\%\\ &=\frac{work\ out}{work\ in}\times 100\%\\ &=\frac{power\ out}{power\ in}\times 100\%\\ \end{aligned}\end{equation} $$ Efficiency can also be written in terms of mechanical advantage and velocity ratio: $$ \begin{equation}\begin{aligned} efficiency=\frac{MA}{VR}\times 100\%\\ \end{aligned}\end{equation} $$
Example
A lever is used to move a box of weight $4\ N$ over a distance of $2\ m$. The force required to do so is $1\ N$ but it has to be applied over a distance of $9\ m$. Find the efficiency of the system. $$ \begin{equation}\begin{aligned} efficiency&=\frac{work\ out}{work\ in}\times 100\%\\ &=\frac{4\ N\times 2\ m}{1\ N\times 9\ m}\times 100\%\\ &=\frac{8\ Nm}{9\ Nm}\times 100\%\\ &=88.9\%\\ \end{aligned}\end{equation} $$ The system has an efficiency less than 100% thus it is not ideal.