## Work

We define work as the product of force on a body and its displacement. In Figure 1 a body undergoes a displacement of $ {\Delta \vec s} = {\vec s _f} - {\vec s _i}$ towards positive x-direction under the action of a constant force ${\vec F}$. But in Figure 2 the body undergoes a displacement of $\Delta \vec s = \vec s_f - \vec s_i$ under the action of a constant force $\vec F$ which makes an angle $\beta$ with the displacement. If $W$ is the work done, we have

\[W = {\vec F} \cdot {\Delta \vec s} \tag{1} \label{1}\]

Note that both ${\vec F}$ and $ {\Delta \vec s} $ are vectors but the work is a scalar quantity, so the work done is a dot (scalar) product of ${\vec F}$ and $ {\Delta \vec s} $. Therefore,

\[W = {F}\Delta s\cos \beta \label{2} \tag{2}\]

Note that friction always does a negative work as it always applies a force in opposite direction of displacement. The friction which does negative work is always kinetic friction. If you consider the friction between the surfaces of the box and the floor, the work done by the constant force $\vec F$ is the maximum work done which is $W_\text{max} = F\Delta s \cos \beta$. The work done by the kinetic friction is $W_k = f_k\Delta s \cos 180 = -f_k\Delta s$. Therefore, the total work done is $W_\text{total} = F\Delta s \cos \beta - f_k\Delta s$ which is the same as the work done by the resultant force. The resultant force is the result of the sum of all forces on a body.

One interesting thing in Figure 2 is that the force $\vec F$ has two components $\vec F_x$ and $\vec F_y$. The sum of the y-component $\vec F_y$ and the normal force is balanced by the weight of the body and hence the normal force decreases. Therefore the kinetic friction decreases due to the y-component of the force $\vec F$. In this case the work is done by the x-component $\vec F_x$ only. Obviously the friction plays negative role here. The unit of work is Joule denoted by J which is $kg \cdot {m^2}/{s^2}$.

If the force is not always constant or the displacement is not always along a straight line, we find the small work through an infinitesimally small displacement (infinitesimally small distance has can be considered as displacement as it has a particular direction) in which the force can be considered as constant. Then we integrate to find the total work done along the whole path. If $dW$ is the small work done by the force $\vec F $ along an infinitesimal displacement $ds$, then the total work done is,

\[W = \int {dW} = \int {Fds\cos \beta } \tag{3}\]

where $\beta$ is the angle between the force and the displacement.

## Power

Power is the measure of how much work is done per unit time or the rate of change of work done. It is the total work done by a force divided by the time interval during the work is done. Let $s$ is the total displacement covered by a body under the action of a constant force of magnitude $F$ in the direction of displacement in time interval $\Delta t = {t_2} - {t_1}$. Then, the power is the total work done divided by the time interval that is,

\[P = \frac{{Fs}}{{{t_2} - {t_1}}} = \frac{W}{{\Delta t}}\]

In case of a variable force or the displacement is not along a straight line we can find the rate of change of work done at any particular instant. Let $dW$ is a very small work done in a very small time interval $dt$ in which $\Delta t$ approaches zero. Note that in a very small time $dt$ the force can be taken as constant and the displacement is also along a straight line, so the power at any instant is,

\[P = \frac{{dW}}{{dt}} \tag{4} \label{4}\]

The unit of power is *watt* denoted by $W$ which is $J/s$ or $kg \cdot {m^2}/{s^3}$. Note that the unit of power and work done denoted by the same letter $W$ are not the same things.