Q.An elevator can carry a maximum load of $1800\ \text{kg}$ (elevator + passengers) is moving up with a constant speed of $2\ \text{m s}^{-1}$. The frictional force opposing the motion is $4000\ \text{N}$. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A person lifts 60 kg load to a vertical height of 30 m over a duration of 20 seconds. If the power of the man is 1323 W, the mass of the man is (A) 30 kg (B) 40 kg (C) 50 kg (D) 60 kg

   ›Reveal solutionSolution

   The man's power output is spent lifting the total weight (himself + the load) through the given height in the given time. Solving $P = (M+60)gh/t$ for $M$ (using $g=9.8\,ms^{-2}$) gives $M=30\,kg$.

   Concept and Intuition

   When a person carries a load up a height, the work done against gravity is on the combined mass — the person's own body plus whatever they're carrying — because both must be raised through the same vertical distance. Average power is just this total work divided by the time taken.

   Step-by-Step Solution

   1. Total work done $= (M + 60)\,g\,h$, where $M$ is the man's mass, $60\,kg$ is the load, $h=30\,m$.
   2. Average power $P = \dfrac{\text{Work}}{\text{time}} = \dfrac{(M+60)gh}{t}$, with $t = 20\,s$.
   3. Substitute the given power $P=1323\,W$ and $g=9.8\,ms^{-2}$:

   $1323 = \frac{(M+60)(9.8)(30)}{20} = (M+60)(14.7)$

   4. Solve: $M + 60 = \dfrac{1323}{14.7} = 90 \Rightarrow M = 30\,kg$.

   Common Mistakes …

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2. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The power required for an engine to maintain a constant speed of $50\ \mathrm{m\,s^{-1}}$ for a train of mass $3\times10^6\ \mathrm{kg}$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is 0.05 and acceleration due to gravity $=10\ \mathrm{m\,s^{-2}}$) (A) 75 MW (B) 40 MW (C) 75 kW (D) 65 MW

   ›Reveal solutionSolution

   This tests that at constant velocity, driving force balances friction, and power is force times speed; the answer is (A).

   Concept and Intuition

   When a vehicle moves at constant speed, its net force is zero, so the engine's tractive force must exactly balance the resistive (friction) force. Once you know that force, the power required to maintain the speed is simply $P=Fv$.

   Step-by-Step Solution

   1. Friction force $=\mu_k mg$, where $\mu_k=0.05$, $m=3\times10^6$ kg, $g=10\ \mathrm{m/s^2}$.
   2. $F=0.05\times3\times10^6\times10=0.05\times3\times10^7=1.5\times10^6$ N.
   3. Since speed is constant, the engine must supply exactly this force to overcome friction. …

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A crane of efficiency 80% is used to lift 8000 kg of coal from a mine of depth 108 m. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in kW) is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) 5 (B) 4 (C) 6 (D) 3

   ›Reveal solutionSolution

   The useful output power (lifting work per unit time) is 2.4 kW; dividing by the 80% efficiency gives the crane's actual power rating, 3 kW.

   Concept and Intuition

   Efficiency $\eta = \dfrac{\text{useful output power}}{\text{power supplied/consumed}}$. The work of lifting the coal against gravity is the useful output; because the crane is only 80% efficient, it must actually consume (draw) more power than this useful output to compensate for losses.

   Step-by-Step Solution

   1. Useful work done $=mgh = 8000\times10\times108 = 8{,}640{,}000\ \mathrm{J}$.
   2. Useful (output) power $=\dfrac{W}{t}=\dfrac{8{,}640{,}000}{3600}=2400\ \mathrm{W}=2.4\ \mathrm{kW}$. …

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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A train of mass $10^6$ kg is moving at a constant speed of 108 kmph. If the frictional force acting on it is 0.5 N per 100 kg, then the power of the train is (A) 300 kW (B) 150 kW (C) 75 kW (D) 225 kW

   ›Reveal solutionSolution

   At constant speed the driving force equals friction; power is force times velocity, giving 150 kW.

   Concept and Intuition

   A train moving at constant speed has zero net force, so the engine's driving force must exactly balance friction. Once that force is known, power delivered is simply $P=Fv$.

   Step-by-Step Solution

   1. Convert speed: $108\text{ km/h}=108\times\dfrac{5}{18}=30\ \mathrm{m/s}$.
   2. Total mass $=10^6$ kg. Friction is given as $0.5$ N per $100$ kg, so total friction: $f=\frac{10^6}{100}\times0.5=10^4\times0.5=5000\ \mathrm{N}.$
   3. At constant velocity, the train's driving force equals this friction force: $F=5000$ N. …

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5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.A circular well of diameter 2 m has water upto the ground level. If the bottom of the well is at a depth of 14 m, the time taken in seconds to empty the well using a 1.4 kW motor is (Acceleration due to gravity $= 10\ \mathrm{m\,s^{-2}}$) (A) 1860 (B) 2200 (C) 2660 (D) 3300

   ›Reveal solutionSolution

   Tests work needed to pump water out of a well against gravity, using energy/power, where different layers of water are lifted through different heights. Answer: 2200 s.

   Concept and Intuition

   Unlike lifting a single mass through a fixed height, emptying a well means water at different depths must be lifted different distances — water near the bottom travels the full 14 m, water near the top travels almost nothing. So we must integrate over depth to get the total work, then use $P = W/t$ to find time, since the motor delivers energy at a constant rate.

   Step-by-Step Solution

   1. Well radius $= 1$ m (diameter 2 m), so cross-sectional area $A = \pi r^2 = \pi\ \mathrm{m^2}$.
   2. Consider a thin layer of water of thickness $dy$ at depth $y$ (measured from the top/ground level, $y$ from 0 to 14 m). Its mass is $dm = \rho A\, dy$, and it must be raised through height $y$ to reach the ground.
   3. Work to raise this layer: $dW = dm\cdot g \cdot y = \rho A g\, y\, dy$.
   4. Total work to empty the well:

   $W = \rho A g \int_0^{14} y\, dy = \rho A g \cdot \frac{14^2}{2} = \rho A g \times 98$

   5. Substituting $\rho=1000\ \mathrm{kg/m^3}$, $A=\pi\ \mathrm{m^2}$, $g=10\ \mathrm{m/s^2}$: …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A motor can pump 7560 kg of water per hour from a well of depth 100 m. If the efficiency of the pump is 70%, then power of the pump is (Acceleration due to gravity $= 10\ \mathrm{m\,s^{-2}}$) (A) 4 kW (B) 6 kW (C) 3 kW (D) 7 kW

   ›Reveal solutionSolution

   Compute the useful gravitational power output, then divide by efficiency to get the pump's actual (input) power: 3 kW.

   Concept and Intuition

   The useful work done by the pump is lifting water against gravity; dividing this useful output rate by the given efficiency gives the actual power the pump consumes/delivers, since efficiency $=\dfrac{\text{output}}{\text{input}}$.

   Step-by-Step Solution

   1. Mass of water pumped per second: $\dfrac{7560\ \text{kg}}{3600\ \text{s}} = 2.1\ \mathrm{kg\,s^{-1}}$.
   2. Useful power output $= \dfrac{mgh}{t} = 2.1\times10\times100 = 2100$ W. …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.A truck of mass 1200 kg moves over an inclined plane raising 1 in 20, with a speed of 18 kmph. The power of the engine is $(g = 10\ \text{ms}^{-2})$ (A) 2 kW (B) 3 kW (C) 3.6 kW (D) 1 kW

   ›Reveal solutionSolution

   The engine must supply a force equal to the gravity component along the incline to keep the truck moving at constant speed; power is simply this force times speed.

   Concept and Intuition

   "Rising 1 in 20" describes the slope as a ratio: for every 20 m travelled along the incline, the truck rises 1 m, so $\sin\theta = 1/20$ (this is the conventional interpretation used for gentle inclines, where $\sin\theta \approx \tan\theta$ for small angles is not even needed — the ratio is $\sin\theta$ by the problem's own convention of rise per unit length along the slope). At constant velocity, net force is zero, so the engine's driving force must exactly balance the component of gravity pulling the truck back down the slope.

   Step-by-Step Solution

   1. Convert speed: $18\ \text{km/h} = \dfrac{18\times1000}{3600} = 5\ \text{m/s}$.
   2. Slope gives $\sin\theta = \dfrac{1}{20}$. …

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8. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.An elevator can carry a maximum load of 2000 kg (elevator + Passengers) is moving up with a constant speed of 9 kmph. If the frictional force opposing the motion is $5 \times 10^3$ N. The minimum power delivered by the motor to the elevator is (Acceleration due to gravity $= 10\ \mathrm{ms^{-2}}$) (A) 26.25 kW (B) 52.5 kW (C) 31.25 kW (D) 62.5 kW

   ›Reveal solutionSolution

   This tests power delivered by a motor against gravity and friction. Both the weight and the frictional resistance must be overcome at constant speed, giving 62.5 kW.

   Concept and Intuition

   For a body moving at constant velocity, the net force is zero, so the driving force from the motor must exactly balance all opposing forces — here, both the elevator's weight (gravity) and the frictional resistance to motion. Power delivered at constant velocity is $P = Fv$, where $F$ is this total balancing force.

   Step-by-Step Solution

   1. Convert speed: $9\ \mathrm{kmph} = \dfrac{9\times1000}{3600} = 2.5\ \mathrm{ms^{-1}}$.
   2. Weight of elevator + passengers: $mg = 2000\times10 = 20000\ \mathrm N$. …

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9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.An object of mass $m$ is projected with an initial velocity $u$ with an angle of $\theta$ with the horizontal. The average power delivered by gravity in reaching the highest point. (A) $\dfrac{mgu\sin^2\theta}{2}$ (B) $\dfrac{mu^2\sin^2\theta}{2g}$ (C) [AMBIGUOUS] (D) $\dfrac{mgu\sin\theta}{2}$

   ›Reveal solutionSolution

   Average power of a constant force equals that force times the average velocity component along it; the vertical velocity drops linearly from $u\sin\theta$ to $0$. Answer: magnitude $\dfrac{mgu\sin\theta}{2}$.

   Concept and Intuition

   For a constant force like gravity, average power over an interval is simply $\vec F\cdot\vec v_{avg}$, because power is linear in velocity and the force doesn't change. The vertical velocity of a projectile decreases uniformly (constant deceleration $g$) from $u\sin\theta$ at launch to $0$ at the top, so its time-average is just the arithmetic mean of the endpoints.

   Step-by-Step Solution

   1. Vertical velocity at launch: $u\sin\theta$; at the highest point: $0$.
   2. Time to reach the top: $t=\dfrac{u\sin\theta}{g}$.
   3. Average vertical velocity (linear decay): $\bar v_y=\dfrac{u\sin\theta+0}{2}=\dfrac{u\sin\theta}{2}$.
   4. Gravity's force is $-mg$ (taking upward positive). Average power $=F\cdot\bar v_y = -mg\cdot\dfrac{u\sin\theta}{2}=-\dfrac{mgu\sin\theta}{2}$. …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The average power generated by a 90 kg mountain climber who climbs a summit of height 600 m in 90 minutes is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) 100 W (B) 25 W (C) 200 W (D) 50 W

    ›Reveal solutionSolution

    Average power is just the total work done (against gravity, climbing to height $h$) divided by the total time taken. Answer: 100 W.

    Concept and Intuition

    Average power over an interval is defined as total work (or energy transferred) divided by total time. Here the "work" is the gravitational potential energy gained in climbing height $h$, and the time must be converted from minutes to seconds to keep units consistent (SI watts).

    Step-by-Step Solution

    1. Work done against gravity: $W = mgh = 90\times10\times600 = 540{,}000\,\mathrm J$.
    2. Convert time: $t = 90\,\text{min} = 90\times60 = 5400\,\mathrm s$.
    3. Average power: $P = \dfrac{W}{t} = \dfrac{540{,}000}{5400} = 100\,\mathrm W$.

    Common Mistakes …

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11. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.The power of an engine which draws 1000 kg of water from a well of depth 9 m in one minute is (Acceleration due to gravity $= 10\ \mathrm{ms^{-2}}$) (A) $1.5\ \mathrm{kW}$ (B) $1.5\ \mathrm{W}$ (C) $150\ \mathrm{W}$ (D) $15\ \mathrm{kW}$

    ›Reveal solutionSolution

    A straightforward work–power calculation for lifting water against gravity. Answer: (A) $1.5\ \mathrm{kW}$.

    Concept and Intuition

    The minimum work required to raise a mass $m$ through height $h$ against gravity is $W = mgh$. Power is the rate of doing this work, $P = W/t$.

    Step-by-Step Solution

    1. Compute work: $W = mgh = 1000\ \mathrm{kg}\times10\ \mathrm{ms^{-2}}\times9\ \mathrm{m} = 90000\ \mathrm{J}$.
    2. Convert time: 1 minute $= 60\ \mathrm{s}$.
    3. Compute power: $P = \dfrac{90000\ \mathrm{J}}{60\ \mathrm{s}} = 1500\ \mathrm{W} = 1.5\ \mathrm{kW}$. …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.A lift raises 50 passengers each having average weight 600 N to a height of 100 m at a constant speed in time T. If the average power of 15 kW is required by the lift, then the value of T in seconds is (A) 150 (B) 100 (C) 300 (D) 200

    ›Reveal solutionSolution

    $T=\text{Work}/\text{Power}=(50\times600\times100)/15000=200$ s.

    Concept and Intuition

    At constant speed, the lift does work against gravity equal to (total weight)$\times$(height). Average power is that total work divided by the time taken, so time is just work divided by power.

    Step-by-Step Solution

    1. Total weight of passengers $=50\times600=30{,}000$ N.
    2. Work done raising them by 100 m: $W=30{,}000\times100=3{,}000{,}000$ J $=3\times10^6$ J.
    3. Average power $P=15$ kW $=15{,}000$ W.
    4. $T=\dfrac{W}{P}=\dfrac{3\times10^6}{15{,}000}=200$ s. …

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