A force is conservative if work done by it is path independent. In other words, the work done by a conservative force on an object moving between two points depends only on the initial and final positions of the object, and not on the specific path taken. The examples of conservative forces are gravitational force, spring force, electrostatic force etc.

Conservative forces $F$ are characterized by having a potential energy function $U$ that depends only on the position of the object i.e., \begin{align} F=-\frac{\mathrm{d}U(x)}{\mathrm{d}x}. \end{align}

If forces acting on a system are conservative then the mechanical energy (sum of kinetic and potential energy) of the system is conserved.

A force which is not conservative is called **non-conservative force**. The work done by a non-conservative force is path dependent. Examples of non-conservative forces include friction, air resistance etc. The mechanical energy of the system is not conserved in the presence of non-conservative forces.

The force on a particle is zero in an equilibrium position. In terms of potential energy $U$ \begin{align} F=-\frac{\mathrm{d}U}{\mathrm{d}x}=0. \end{align}

The second derivative of the potential energy gives type of the equilibrium. If the second derivative $\mathrm{d}^2U/\mathrm{d}x^2 < 0$ then equilibrium is stable.

If the second derivative $\mathrm{d}^2U/\mathrm{d}x^2 > 0$ then equilibrium is unstable.

If the second derivative $\mathrm{d}^2U/\mathrm{d}x^2 = 0$ then equilibrium is neutral (or metastable). The type of equilibrium can be easily found out from the potential energy curve.

**Problem (IIT JEE 2018):**
A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $x$-axis. Its kinetic energy $K$ changes with time as $\mathrm{d}K/\mathrm{d}t=\gamma t$, where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

- The force applied on the particle is constant.
- The speed of the particle is proportional to time.
- The distance of the particle from the origin increases linearly with time.
- The force is conservative.

**Solution:**
The kinetic energy of a particle of mass $m$ moving with a speed $v$ is given by $K=\frac{1}{2}mv^2$. Differentiate $K$ with time to get
\begin{align}
\frac{\mathrm{d}K}{\mathrm{d}t}&=mv\frac{\mathrm{d}v}{\mathrm{d}t} \\
&=\gamma t.
\end{align}
Integrate with initial condition $v=0$ at $t=0$ (because the particle is initially at rest)
\begin{align}
m\int_0^v v\, \mathrm{d}v=\gamma\int_0^t t\,\mathrm{d}t, \nonumber
\end{align}
to get
\begin{align}
v=\sqrt{\frac{\gamma}{m}}\; t.
\end{align}
Thus, speed of the particle is proportional to time. Integrate above equation with initial condition $x=0$ at $t=0$ (because the particle is initially at the origin)
\begin{align}
\int_0^x \mathrm{d}x=\sqrt{\frac{\gamma}{m}} \int_0^t t\,\mathrm{d}t, \nonumber
\end{align}
to get
\begin{align}
x=\frac{1}{2}\sqrt{\frac{\gamma}{m}}\; t^2. \nonumber
\end{align}
Hence, distance of the particle from the origin increases quadratically (not linearly) with time. Use the above equation to get the acceleration of the particle
\begin{align}
a &=\frac{\mathrm{d}v}{\mathrm{d}t} \\
&=\sqrt{\frac{\gamma}{m}}. \nonumber
\end{align}
Apply Newton's second law to get the applied force on the particle
\begin{align}
F=ma=\sqrt{\gamma m}. \nonumber
\end{align}
Thus, force applied on the particle is constant.

What about the conservative nature of the force? A force is conservative if work done by the force in any closed loop is zero (i.e., work done is independent of the path). If force is constant (both magnitude and direction) then work done by the force in a closed loop is given by \begin{align} W&=\oint \vec{F}\cdot\mathrm{d}\vec{r} \\ &=\vec{F}\cdot\oint\mathrm{d}\vec{r} \\ &=\vec{F}\cdot\vec{0}\\ &=0. \nonumber \end{align} Note that we are able to take $\vec{F}$ out of integral sign because $\vec{F}$ is constant and $\oint\mathrm{d}\vec{r}$ is zero because integration is carried out over a closed loop. The argument `since force is constant, hence it is conservative' should be understood properly. The kinetic friction force and viscous drag force (when body moves with a terminal speed) are both constant (in magnitude) but are non-conservative. These forces cannot start the particle motion from the rest. The situation given in this problem is similar to a particle dropped from rest under the gravitational force. This force is constant, can initiate the motion given in the problem and is conservative.

**Problem:**
A block of mass $m$ slides along a horizontal table with speed $v_0$. At $x=0$ it hits a spring of spring constant $k$ and begin to experience a frictional force. The coefficient of friction is variable and is given by $\mu=bx$, where $b$ is a constant. The loss in mechanical energy when the block comes to momentarily rest for the first time is,

**Solution:**
Let the block comes to rest after compressing the spring by a distance $x=l$. The frictional force on the block is $f=\mu mg=b mgx$. The work done by the frictional force in displacing the block by a displacement $l$ is,
\begin{align}
W_f & =-\int_0^{l}\!f\,\mathrm{d}x \\
&=-mgb\int_0^{l}\! x\,\mathrm{d}x \\
&=-\tfrac{1}{2}mgbl^2.\nonumber
\end{align}
The initial kinetic energy of the block is $K_i=\tfrac{1}{2}mv_0^2$, and the final kinetic energy (when the block comes to momentary rest) is $K_f=0$. The initial potential energy of the spring is $U_i=0$, and its final potential energy is $U_f=\tfrac{1}{2}kl^2$.

Consider a system taken from an initial state $i$ to a final state $f$ by conservative and non-conservative forces. The work done by conservative forces is equal to the increase in the system's potential energy i.e., $W_{i\to f}^{c}=U_i-U_f$ (e.g., the work done by the spring force when it is compressed from $x=0$ to $x=l$ is $\tfrac{1}{2}kl^2$). The work done by non-conservative forces is denoted by $W_{i\to f}^\text{nc}$. A nice way to write the work energy theorem, $K_f-K_i=W_{i\to f}^{c}+W_{i\to f}^{nc}$, is
\begin{align}
(K_f+U_f)-(K_i+U_i)=W_\mathrm{i\to f}^\text{nc}.\nonumber
\end{align}
*The change in mechanical energy is equal to the work done by the non-conservative forces.* Substitute values in the above equation,
\begin{align}
\left(0+\tfrac{1}{2}kl^2\right)-\left(\tfrac{1}{2}mv_0^2+0\right)=-\tfrac{1}{2}mgbl^2,\nonumber
\end{align}
and simplify to get
\begin{align}
l^2={\frac{mv_0^2}{k+mgb}}.\nonumber
\end{align}
The loss in mechanical energy is,
\begin{align}
\Delta E&=(K_i+U_i)-(K_f+U_f) \\
&=\tfrac{1}{2}mgbl^2\nonumber\\
&=\frac{mgb}{k+mgb}\left(\tfrac{1}{2}mv_0^2\right).\nonumber
\end{align}