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Work, Energy and Power
Power
11 questions from 10 papers, each with its answer.
- questions
- 11
- papers
- 10
- years asked, 2017–2026
- 2 of 10
- of Work, Energy and Power
- 15.49%
Appears in
Concepts tested
- Average power
- Constant power motion
- energy = power × time
- Instantaneous power
- Motion under constant power
- Power as dot product
- power delivered by a lift motor
- Power from falling water
- Power of a pump
- power output of a turbine
- Rate of kinetic energy
All Questions
2022 2 questions
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An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of $1.5\ ms^{-1}$. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : ($g=10\ ms^{-2}$)The lift moves with constant velocity, so a = 0. Tension (force by motor) $T = W + f = 2000 \times 10 + 3000 = 23000$ N Power $= Tv = 23000 \times 1.5 = 34500$ W
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The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :$E = P \times t = 100 \times 10^3 \times 3600$ $= 36 \times 10^7$ J
2021 1 question
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Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? ($g=10\ m/s^2$)$P_{in} = \frac{mgh}{t} = 15 \times 10 \times 60 = 9000$ W Losses are 10%, so $P_{out} = 90\%$ of $P_{in} = 0.9 \times 9000 = 8100$ W = 8.1 kW
2015 AIPMT-I 1 question
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A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t isWork done in time t: $W = Pt = kt$
By the work-energy theorem, $\frac{1}{2}mv^2 = kt \Rightarrow v = \sqrt{\frac{2kt}{m}}$
$P = Fv \Rightarrow F = \frac{k}{v} = k\sqrt{\frac{m}{2kt}} = \sqrt{\frac{mk}{2t}}$
$F = \sqrt{\frac{mk}{2}}\,t^{-1/2}$
2012 AIPMT-MAINS 1 question
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A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P_0$. The instantaneous velocity of this car is proportional to$P_0 = Fv = m\frac{dv}{dt}v$
$v\,dv = \frac{P_0}{m}dt$
Integrating from rest: $\frac{v^2}{2} = \frac{P_0}{m}t$
$v = \sqrt{\frac{2P_0t}{m}} \Rightarrow v \propto t^{1/2}$
2011 AIPMT-PRE 1 question
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A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatestPower delivered by a force is $P = \vec{F}\cdot\vec{v} = Fv\cos\theta$
Just after projection, the velocity is upward and gravity is downward, so $\theta = 180^\circ$ and the power is negative.
At the highest point, v = 0, so the power is zero.
Just before the body hits the earth, the velocity is downward, i.e., along the force ($\theta = 0^\circ$, $\cos\theta = 1$).
At that instant the speed is maximum and the gravitational force is also maximum (the body is nearest to the earth), so $P = Fv$ is the greatest.
Hence the power is greatest at the instant just before the body hits the earth.
2010 AIPMT-MAINS 1 question
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A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is V, the power delivered to the particle isBy the work–energy theorem, work done on the particle = gain in kinetic energy: $W = \frac{1}{2}MV^2 - 0 = \frac{1}{2}MV^2$
This work is delivered in time T.
Power delivered (average) = $\frac{W}{T}$
$P = \frac{1}{2}\frac{MV^2}{T}$
2010 AIPMT-PRE 1 question
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An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?Mass of water flowing out per second: $\frac{dm}{dt} = (\text{mass per unit length})\times v = 100\times2 = 200$ kg/s
Force exerted by the engine on the water = rate of change of momentum = $v\frac{dm}{dt} = 2\times200 = 400$ N
Power = force × velocity = $Fv = 400\times2$
In symbols, with $\lambda$ = mass per unit length: $P = \lambda v^3 = 100\times2^3$
P = 800 W
2009 AIPMT 1 question
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An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?In time dt, a length $v\,dt$ of the water jet leaves the hose.
Mass of water leaving in time dt: $dM = m\times v\,dt$, so $\frac{dM}{dt} = mv$
Kinetic energy carried by this mass: $dK = \frac{1}{2}(dM)v^2$
Rate at which kinetic energy is imparted: $\frac{dK}{dt} = \frac{1}{2}\frac{dM}{dt}v^2 = \frac{1}{2}(mv)v^2$
$\frac{dK}{dt} = \frac{1}{2}mv^3$
2008 AIPMT 1 question
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Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10 % of energy. How much power is generated by the turbine ? ($g = 10$ m/s$^2$)Energy available per second to operate the turbine $= \dfrac{m}{t}\,g\,h$
$= 15 \times 10 \times 60 = 9000$ joule per second
Power supplied to the turbine = 9000 W
Power lost due to friction = 10 percent of 9000 = 900 W
Power generated by the turbine $= 9000 - 900 = 8100$ W
$= 8.1$ kW
Second approach: $P_{generated} = P_{input} \times \dfrac{90}{100} = \dfrac{mgh}{t} \times \dfrac{90}{100} = 15 \times 10 \times 60 \times \dfrac{90}{100} = 8.1$ kW
2000 AIPMT 1 question
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If $\vec{F} = (60\hat{i} + 15\hat{j} - 3\hat{k})$ N and $\vec{V} = (2\hat{i} - 4\hat{j} + 5\hat{k})$ m/s, then instantaneous power is :Instantaneous power: $P = \vec{F}\cdot\vec{v}$
$P = (60\hat{i} + 15\hat{j} - 3\hat{k})\cdot(2\hat{i} - 4\hat{j} + 5\hat{k})$
$= (60)(2) + (15)(-4) + (-3)(5) = 120 - 60 - 15$
$P = 45$ watt