Power
By Power is the rate of doing work i.e., \begin{align} P=\frac{W}{t}. \end{align} Its unit is Watt and dimensions are ML2T-3. The power $P$ delivered by a force $\vec{F}$ is related to the velocity $\vec{v}$ by \begin{align} P=\vec{F}\cdot\vec{v}. \end{align}
Problems from IIT JEE
Problem (IIT JEE 2013): A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed of the particle is zero, the speed after 5 s is_______ m/s.
Solution: The power $P$ delivered by a force $\vec{F}$ is related to velocity $\vec{v}$ by \begin{align} P=\vec{F}\cdot\vec{v}. \end{align} From Newton's second law, \begin{align} \vec{F}=m\;{\mathrm{d}\vec{v}}/{\mathrm{d}t}. \end{align} In one-dimension, above equations give \begin{align} P=mv\;{\mathrm{d}v}/{\mathrm{d}t}. \end{align} Integrate with time ($P$ is constant) to get \begin{align} v^2=u^2+{2Pt}/{m}.\nonumber \end{align} Substitute the values of $u$, $P$, $t$ and $m$ to get $v=5$ m/s.
Problem (JEE Mains 2020): A 60 HP electric motor lifts an elevator having a maximum load capacity of 2000 kg. If the frictional force of the elevator is 4000 N, the speed of the elevator at full load is close to (1 HP = 746 W, g=10 m/s2)
- 1.5 m/s
- 1.7 m/s
- 2.0 m/s
- 1.9 m/s
Solution: The net force on the elevator at the full load is, \begin{align} F & =m_\text{max}g+f \\ &=(2000)(10)+4000 \\ &=24000\;\mathrm{N}.\nonumber \end{align} Let $v$ be the elevator's speed at the full load. The power delivered to the elevator is $P_\text{elevator}=Fv$. This power is delivered by an electric motor of power $P_\text{motor}=60$ HP. At the full load, $P_\text{elevator}=P_\text{motor}$. Substitute values to get, \begin{align} v&=\frac{P_\text{motor}}{F}\\ &=\frac{60\times746}{24000}\\ &=1.9\;\mathrm{m/s}.\nonumber \end{align}
Problem (IIT JEE 2000): A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electric energy. For wind speed $v$, the electric power output will be proportional to,
- $v$
- $v^2$
- $v^3$
- $v^4$
Solution: Let $\rho$ be the density of the air and $A$ be the cross-section area of the blades. Consider a cylindrical volume $V$ with cross-sectional area $A$ and length $x$ (see figure).
The mass of air contained in volume $V$ is $m=\rho A x$ and its kinetic energy is, \begin{align} K&=\frac{1}{2}mv^2\\ &=\frac{1}{2}\rho A x v^2.\nonumber \end{align} The time taken by this volume to intersect the blades is $\Delta t=x/v$. The air transfers its entire kinetic energy to the blades (assuming $100\; \%$ efficiency). Thus, the power generated is, \begin{align} P&=\frac{K}{\Delta t}\\ &=\frac{1}{2}\rho A v^3.\nonumber \end{align}
Problem (IIT JEE 1984): A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to
- $t^{1/2}$
- $t^{3/4}$
- $t^{3/2}$
- $t^{2}$
Solution: The power of a force $F$ acting on a body of mass $m$ moving with a velocity $v$ is given by \begin{align} P &=Fv \\ &=m\left(\mathrm{d}v/\mathrm{d}t\right) v. \end{align} Integrate with initial condition $v=0$ at $t=0$ to get \begin{align} v&=\mathrm{d}x/\mathrm{d}t \\ &=(2Pt/m)^{1/2}. \end{align} Integrate to get \begin{align} x=\sqrt{\frac{8P}{9m}}\; t^{3/2}. \end{align}
