# Work-Energy Theorem

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.

## Problems from IIT JEE

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

1. 4.5 J
2. 22 J
3. 9 J
4. 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=3m/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}