The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy i.e., \begin{align} W=\Delta K=K_f-K_i, \end{align} where $K_i$ and $K_f$ are the initial and the final kinetic energies. The work energy theorem is a useful tool for solving problems. This theorem is applied for work done by conservative forces. It does not take into account non-conservative forces such as friction, which can lead to energy loss.

**Problem (JEE Mains 2017):**
A time-dependent force $F=6t$ acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 s will be

- 4.5 J
- 22 J
- 9 J
- 18 J

**Solution:**
Newton's second law gives particle's acceleration,
\begin{align}
\frac{\mathrm{d}v}{\mathrm{d}t}&=\frac{F}{m}\\
&=\frac{6t}{1}\\
&=6t.\nonumber
\end{align}
Integrate,
\begin{align}
\int_{0}^{v}\!\mathrm{d}v=\int_0^t\!6t\,\mathrm{d}t,\nonumber
\end{align}
to get the velocity $v=3t^2$. The particle's velocity at $t=1$ s$ is $v=3(1)^2=3$ m/s and its kinetic energy is,
\begin{align}
K&=\frac{1}{2}mv^2 \\
&=\frac{1}{2} (1) (3)^2 \\
&=4.5\;\mathrm{J}.\nonumber
\end{align}
By work-energy theorem, the work done by the force in the first 1 s is,
\begin{align}
W &=\Delta K \\
&=K_f-K_i \\
&=4.5-0 \\
&=4.5\;\mathrm{J}.\nonumber
\end{align}

Problem (IIT JEE 2014): Consider an elliptically shaped rail PQ in the vertical plane with $\mathrm{OP}={3}\;\mathrm{m}$ and $\mathrm{OQ}={4}\;\mathrm{m}$. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see figure). Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n\times10)$ Joules. The value of $n$ is (Given $g={10}\;\mathrm{m/s^2}$.),

**Solution:**
Consider a small displacement $\mathrm{d}\vec{s}$ along the elliptical path. Resolve $\mathrm{d}\vec{s}$ in directions parallel and perpendicular to $\vec{F}$, say $\mathrm{d}\vec{s}_{\parallel }$ and $\mathrm{d}\vec{s}_{\perp}$. The work done by $\vec{F}$ for the path from P to Q is,
\begin{align}
W &=\int\vec{F}\cdot\mathrm{d}\vec{s} \\
&=\int\vec{F}\cdot\mathrm{d}\vec{s}_{\parallel}\\
&= F\int\mathrm{d}\vec{s}_{\parallel}\\
&=18(5)={90}\;\mathrm{J}.
\end{align}
By work-energy theorem,
\begin{align}
W=\Delta K+\Delta U,
\end{align}
where $\Delta U=U_f-U_i$ is the change in the potential energy and $\Delta K$ is the change in the kinetic energy. From above equations,
\begin{align}
\Delta K&=W-mgh\\
&=90-1(10)4\\
&={50}\;\mathrm{J}.\nonumber
\end{align}