Work-Energy Theorem
By 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
- 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}
