Consider a spring-mass system. Let us assume a mass, m lying on a smooth horizontal table as shown in the figure. Here, x = 0 is the equilibrium position. One end of the spring is attached to a rigid wall and the other end to the mass.
As long as the spring remains in equilibrium position, its potential energy is zero. Now an, external force \(\bar{F},\) is applied 'so that it is stretched by a distance (x) in the direction of the force.
There is a restoring force called spring force Fs developed in the spring which tries to bring the mass back to its original position. This applied force and the spring force are equal in magnitude but opposite in direction i.e., \(\bar{F}_a=-\bar{F}_s.\) 

According to Hooke's law, the, restoring force developed in the spring is
\(\bar{F}_s=-k.\bar{x}\)    ...(1)
The negative sign 'in the above expression implies that the spring force is always opposite to that of displacement i and k is the force constant. Therefore applied force is \(\bar{F}_a=+k.\bar{x}\). The positive sign implies that the applied force is in the direction of displacement \(\bar{x}\). The spring force is an example of variable force as it depends on the displacement \(\bar{x}\). Let the spring be stretched to a small distance d \(\bar{x}\). The work done by the applied force on the spring to stretch it by a displacement x is stored as elastic potential energy.
\(U=\int{\bar{F}_ad\bar{r}}=\int_{0}^{\pi}|\bar{F a}||d\bar{r}|\cos\theta\)
\(U=\int_{0}^{x}F_adx \cos \theta\)   ...(2)
The applied force \(\bar{F}_a\) and the displacement \(d\bar{r}\) (i.e., here dx) are in the same direction. As, the initial position is taken as the equilibrium position or mean position, x = 0 is the lower limit of integration.
\(U=\int_{0}^{\pi}kxdx\)    ........(3)
\(U=k{\left[ {x^2 \over 2} \right]}_{x}^{0}\) ............(4)
\(U={1\over2}kx^2\)    ........(5)
If the initial position is not zero, and if the mass is changed from position x_i to x_f, then the elastic potential energy is
\(U={1\over2}k({x}_{f}^{2}-{x}_{i}^{2})\)  .......(6)
From equations (5) and (6), we observe that the potential energy of the stretched spring depends on the force constant k and elongation or compression x.

The gravitational potential energy (U) at some height his, equal to the amount of work required to take the object from the ground to that height h.
\(U=\int{\bar{F}_a.d\bar{r}}=\int_{0}^{h}|\bar{F}_a||d\bar{r}|\cos\ \theta\)
Since the displacement and the applied force are in the same upward direction, the angle between them, \(\theta = 0°.\)Hence, cos°=1 and \(|\bar{F}_a|=mg\) and \(|d\bar{r}|=dr.\)
\(U=mg\int_{0}^{4}dr\Rightarrow mg {[r]}_{0}^{h} =mgh\)

Consider an object of mass m moving with a velocity v. Then its linear momentum is P = mv and its kinetic energy, \(KE={1\over2}{mv}^{2}\)
\(KE={1\over2}{mv}^{2}={1\over 2}m(\bar{v}.\bar{v})\)   ...(1)
Multiplying both the numerator and denominator of equation (1) by mass, m
\(KE={1\over2}{m^2(\bar{v}.\bar{v})\over m}={1\over2}{(m\bar{v}).(m\bar{v})\over m}\)       \([\bar{p}=m\bar{v}]\)
\(={1\over2}{\bar{p}.\bar{p}\over m}={p^2\over 2m}\)
\(\mathrm{KE}=\frac{p^{2}}{2 m}\)   ....(2)
where \(|\vec{p}|\) is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by
\(|\vec{p}|=\mathrm{p}=\sqrt{2 m(K \cdot E)}\)  ....(3)

(i) If the work done by the force on the body is positive then its kinetic energy increases.
(ii) If the work done by the force on the body is negative then its kinetic energy decreases.
(iii) If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

Positive work: If a force acting on a body has a component in a direction of the displacement.
Eg: When a body falls under free fall. (8 = 0°)
Negative work: If a force acting on a body has a component in the opposite direction of the displacement.  
Eg: When brakes are applied to a moving vehicle, the work done by the brake force is negative.
Zero work: If the body gets displaced along a direction perpendicular to the direction of applied force. Eg: When a coolie walks on a horizontal platform with load.