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A lift motor, a pump filling a rooftop tank and a car on a highway all raise the same exam question: how much power does it need? These problems are short, but they mix units (minutes, litres, kilowatt-hours, horsepower) and an efficiency that students often apply the wrong way round. This post gives you the two formulas, three worked examples and a check for each.
What are the formulas for power?
NCERT's Work, Energy and Power chapter defines average power as work divided by time, P = W/t, and instantaneous power as P = F·v, the force dotted with the velocity of the point where it acts. The SI unit is the watt, 1 W = 1 J/s, and NCERT gives the older unit 1 hp = 746 W, still used for vehicles and pumps. OpenStax University Physics gives the same definitions and conversion.
Efficiency compares what you get with what you put in:
η = useful output power / input power
Input power = useful output power / η
Because η is less than 1, the input is always larger than the output. If your input comes out smaller, you have divided the wrong way.
The JEE (Main) 2026 syllabus lists power in Unit 4, Work, Energy and Power. Check your exam year's bulletin for changes.
How much power does a lift motor need?
At constant speed, the motor's pull must balance the weight plus any friction. Then P = Fv.
Example 1. A lift and its passengers have a total mass of 1200 kg. It rises at a steady 1.5 m/s against a friction force of 3000 N. Take g = 10 m/s². Find the power the motor delivers, and the electrical input if the motor is 80% efficient.
Force needed: F = mg + friction = 12 000 + 3000 = 15 000 N.
Power delivered: P = Fv = 15 000 × 1.5 = 22 500 W = 22.5 kW.
In horsepower: 22 500 / 746 ≈ 30.2 hp.
Electrical input: 22 500 / 0.8 = 28 125 W ≈ 28.1 kW.
NCERT's own lift example works the same way, with an 1800 kg load at 2 m/s against 4000 N of friction, giving 44 000 W or about 59 hp.
If the lift is accelerating, the pull is larger than the weight plus friction, so the power changes from moment to moment. Find F from F − mg − friction = ma, then use P = Fv at the speed asked about.
How do you size a water pump?
A pump lifts water against gravity and also gives it speed at the outlet. Both count as useful work.
Example 2. A pump fills a rooftop tank 15 m above it with 1000 litres of water (1000 kg) in 10 minutes. The water leaves the pipe at 2 m/s. The pump set is 60% efficient. Find the input power and the energy used in kilowatt-hours.
Work against gravity: mgh = 1000 × 10 × 15 = 150 000 J.
Kinetic energy at the outlet: ½ × 1000 × 2² = 2000 J.
Useful power: (150 000 + 2000) / 600 s ≈ 253 W. Note 10 minutes = 600 s.
Input power: 253.3 / 0.6 ≈ 422 W.
Energy used: 422.2 W × 600 s ≈ 253 000 J. Since 1 kWh = 3.6 × 10⁶ J, that is about 0.070 kWh.
Many textbook questions ignore the outlet speed; here it adds just over 1%. When the water is fired out fast, as in a fountain or a fire hose, it can dominate.
What is the top speed at constant power?
A vehicle with a fixed engine power P speeds up until the driving force P/v falls to the size of the resistance. Then it cannot go any faster.
Example 3. A 1000 kg car's engine delivers 30 kW. The total resistance is a constant 1500 N.
Top speed: v = P/F = 30 000 / 1500 = 20 m/s, which is 72 km/h.
Acceleration at 10 m/s: driving force = 30 000 / 10 = 3000 N, so a = (3000 − 1500) / 1000 = 1.5 m/s².
This explains why a car accelerates hard at low speed and gently near its top speed, even with the same engine power.
What mistakes cost marks in power problems?
Leaving time in minutes or hours while the work is in joules.
Forgetting friction in lift problems, or the outlet speed when it is given.
Inverting efficiency, so the input comes out smaller than the output.
Using the net force in P = Fv instead of the applied force.
Mixing kWh and J. A 1000 W device running for one hour uses 1 kWh, which is 3.6 MJ.
The work in these problems is the same work you meet in the work-energy theorem; power just divides it by time.
A practice routine for power and efficiency
Write every given quantity in SI units in a list before you start.
Solve two lift problems: one at constant speed, one accelerating.
Solve two pump problems: one with and one without outlet speed.
Solve two constant-power vehicle problems asking for top speed and acceleration.
Check every input-power answer is larger than the output.
If you keep choosing the wrong force for P = Fv, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, work out the power a pump needs to fill your own building's tank, using a height and tank size you can measure.
