1. Physics
  2. Work, Energy and Power
  3. Power

Work, Energy and Power · JEE & NEET Physics

Power: notes and previous year questions

The rate of doing work: P = W/t and P = F · v, units and the kilowatt hour, lifting, top speed, constant power, and power–time graphs.

Power in short

  • Power is the rate of doing work or of moving energy.
  • Average power is the total work divided by the total time.
  • Instantaneous power is F · v: the force along the motion times the speed.
  • 1 W = 1 J/s; 1 hp = 746 W; the kilowatt hour is a unit of energy.

1How fast work is done

Two lifts raise identical 10 kg boxes 2 m. Both do mgh=200mgh = 200 J of work, but one takes 2 s and the other 8 s. The fast one delivers 100 W, the slow one 25 W. Power tells you how fast work is done, or energy is moved.

P=dWdtP = \frac{dW}{dt}
Pavg=Wtotalttotal=ΔEΔtP_{\text{avg}} = \frac{W_{\text{total}}}{t_{\text{total}}} = \frac{\Delta E}{\Delta t}

The unit is the watt: 1 W = 1 J/s. For example, 1000 J in 5 s is 200 W, and a 50 kg student running 3 m up the stairs in 5 s delivers 1500/5=3001500/5 = 300 W.

2Instantaneous power: P = F · v

In a tiny time dtdt a force does work F⃗⋅dr⃗\vec F\cdot d\vec r. Dividing by dtdt:

P=F⃗⋅v⃗=Fvcos⁡θP = \vec F\cdot\vec v = Fv\cos\thetaθ is the angle between the force and the velocity: the part of the force along the motion, times the speed.
  • Towing a cart at a steady 4 m/s with 50 N at 60°: P=50×4×cos⁡60∘=100P = 50 \times 4 \times \cos 60^\circ = 100 W.
  • F⃗=2i^+3j^\vec F = 2\hat i + 3\hat j N on a body moving at v⃗=4i^+j^\vec v = 4\hat i + \hat j m/s: P=8+3=11P = 8 + 3 = 11 W.
  • A force at 90° to the velocity, like a string whirling a stone at steady speed, delivers no power.

3Units of power

UnitValueMeasures
watt (W)1 J/s = 1 kg m² s⁻³power
kilowatt (kW)1000 Wpower
megawatt (MW)10610^6 Wpower
horsepower (hp)746 Wpower
kilowatt hour (kWh)3.6×1063.6 \times 10^6 Jenergy

4Lifting and pumping

Lifting at a steady speed, the pull equals the weight, so P=FvP = Fv becomes:

P=mgvP = mgv
  • A crane lifting 500 kg at 2 m/s: P=5000×2=10P = 5000 \times 2 = 10 kW. A 6 kW motor can lift 300 kg at 6000/3000=26000/3000 = 2 m/s.
  • A pump lifting 100 kg of water 20 m in 50 s: W=20 000W = 20\,000 J, P=400P = 400 W.
  • A pump raising 60 kg of water per minute to 10 m: 60006000 J per 60 s = 100 W.

5Vehicles and top speed

An engine delivering power PP pushes with force P/vP/v. At low speed that force is large and the car speeds up; as vv grows the force shrinks, until it just balances the resistance ff. Then the car stops speeding up: that is its top speed.

P=fv,vmax⁡=PfP = fv, \qquad v_{\max} = \frac{P}{f}On a level road at steady speed.
P=(mgsin⁡θ+f) vP = (mg\sin\theta + f)\,vUp a slope at steady speed.
f=kv2  ⇒  P=kv3, vmax⁡=(P/k)1/3f = kv^2 \;\Rightarrow\; P = kv^3,\ v_{\max} = (P/k)^{1/3}

6Motion under constant power

Constant power does not mean constant force: F=P/vF = P/v falls as the speed grows. From P=mv dv/dtP = mv\,dv/dt, P dt=mv dvP\,dt = mv\,dv, and from rest Pt=12mv2Pt = \tfrac{1}{2}mv^2 (all the work becomes kinetic energy).

v=2Ptmv = \sqrt{\frac{2Pt}{m}}v grows as √t.
x=232Pm t3/2x = \frac{2}{3}\sqrt{\frac{2P}{m}}\,t^{3/2}x grows as t^(3/2).

Writing dv/dt=v dv/dxdv/dt = v\,dv/dx instead gives P=mv2 dv/dxP = mv^2\,dv/dx, so Px=13mv3Px = \tfrac{1}{3}mv^3:

v3=3Pxmv^3 = \frac{3Px}{m}

7Power–time graphs

W=∫P dtW = \int P\,dtThe work done is the area under the power–time graph. For constant power, W = Pt.

Example: the power rises steadily to 200 W over 4 s, then stays at 200 W. By 4 s the triangle gives 400 J; each further second adds 200 J, so the work reaches 1000 J at 7 s and 1200 J at 8 s, an average of 150 W.

A 1000 kg car going from rest to 20 m/s in 10 s gains 200 kJ of kinetic energy: an average power of 20 kW (ignoring friction).

Summary

Key ideas

  • Power is the rate of doing work or of moving energy.
  • Average power is the total work divided by the total time.
  • Instantaneous power is F · v: the force along the motion times the speed.
  • 1 W = 1 J/s; 1 hp = 746 W; the kilowatt hour is a unit of energy.
  • Lifting at a steady speed needs P = mgv.
  • A vehicle reaches top speed when its driving force P/v equals the resistance.
  • Under constant power from rest, v grows as √t and x as t^(3/2).
  • The work done is the area under the power–time graph.

Every equation

Power
P=dWdtP = \frac{dW}{dt}
Average power
Pavg=WtP_{\text{avg}} = \frac{W}{t}
Instantaneous power
P=F⃗⋅v⃗=Fvcos⁡θP = \vec F\cdot\vec v = Fv\cos\theta
Watt
1 W=1 J/s1\ \text{W} = 1\ \text{J/s}
Horsepower
1 hp=746 W1\ \text{hp} = 746\ \text{W}
Kilowatt hour
1 kWh=3.6×106 J1\ \text{kWh} = 3.6 \times 10^6\ \text{J}
Lifting at steady speed
P=mgvP = mgv
Level road
P=fvP = fv
Top speed
vmax⁡=P/fv_{\max} = P/f
Up a slope
P=(mgsin⁡θ+f)vP = (mg\sin\theta + f)v
Drag ∝ v²
vmax⁡=(P/k)1/3v_{\max} = (P/k)^{1/3}
Constant power: speed
v=2Pt/mv = \sqrt{2Pt/m}
Constant power: distance
x=232P/m  t3/2x = \tfrac{2}{3}\sqrt{2P/m}\;t^{3/2}
Constant power: speed and distance
v3=3Px/mv^3 = 3Px/m
Work from a P–t graph
W=∫P dtW = \int P\,dt

Previous year questions with solutions

Real JEE and NEET questions on power. Try each one before you open the solution.

Q1NEET 2026One correct option

The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: (g=9.8m/s2)(g=9.8 m/s^{2})

  1. A19.6 W
  2. B39.2 W
  3. C19.6 kW
  4. D39.2 kW
Show answer and solution

Answer: Option C

The crane's work is lifting the load: mgh=1000×9.8×20=1.96×105 Jmgh = 1000 \times 9.8 \times 20 = 1.96 \times 10^{5}\ \mathrm{J}. Spread over 10 s10\ \mathrm{s}, P=1.96×10510=1.96×104 W=19.6 kWP = \dfrac{1.96 \times 10^{5}}{10} = 1.96 \times 10^{4}\ \mathrm{W} = 19.6\ \mathrm{kW}. Two traps sit in the options: 19.6 W19.6\ \mathrm{W} is the right digits with the factor of 10001000 lost, and 39.239.2 comes from dividing by 55 instead of 1010 or from doubling the height. A thousand kilograms raised twenty metres in ten seconds is serious machinery — kilowatts, not the watts of a light bulb.

Q2JEE Main 2025One correct option

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., dmdt∝v\frac{dm}{dt}\propto \sqrt{v}. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

  1. AP ∝v\propto \sqrt{v}
  2. BP ∝v\propto v
  3. CP2∝v3P^{2}\propto v^{3}
  4. DP2∝v5P^{2}\propto v^{5}
Show answer and solution

Answer: Option D

At constant belt speed the force needed is F=vdmdtF = v\dfrac{dm}{dt}, so the power is P=Fv=v2dmdtP = Fv = v^{2}\dfrac{dm}{dt}. Here dmdt=kv\dfrac{dm}{dt} = k\sqrt{v}, so P=kv2v=kv5/2P = kv^{2}\sqrt{v} = kv^{5/2}, and squaring, P2=k2v5P^{2} = k^{2}v^{5}, so P2∝v5P^{2} \propto v^{5}. Option A keeps only the sand rate and forgets that both the force and the speed carry a factor of vv. C is what you get from a single factor of vv: using F=vdmdtF = v\dfrac{dm}{dt} as if it were the power.

Q3JEE Advanced 1994One correct option

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration aca_{c} is varying with time t as aca_{c} = k²rt² where k is a constant. The power delivered to the particle by the force acting on it is:

  1. A2πmk2r2t2\pi mk^{2}r^{2}t
  2. Bmk2r2tmk^{2}r^{2}t
  3. C(mk4r2t5)3\frac{(mk^{4}r^{2}t^{5})}{3}
  4. Dzero
Show answer and solution

Answer: Option B

You have seen this one before, in circular motion; now you know why the method works. F⃗⋅v⃗\vec{F}\cdot\vec{v} kills the inward part of the force, so only the tangential part delivers power. From v2r=k2rt2\dfrac{v^{2}}{r} = k^{2}rt^{2}, v=krtv = krt; then at=dvdt=kra_{t} = \dfrac{dv}{dt} = kr, Ft=mkrF_{t} = mkr, and P=Ftv=(mkr)(krt)=mk2r2tP = F_{t}v = (mkr)(krt) = mk^{2}r^{2}t. D would be right for the centripetal force alone, but the particle is plainly speeding up, and something is paying for that.

Practice questions, easy to hard

Three questions from the power practice ladder: one easy, one medium, one hard.

Q4One correct option

Engines and motors are still often rated in an older unit, the horsepower: 1 hp=746 W1\ \mathrm{hp} = 746\ \mathrm{W}. It is a unit of power, not of work, so a rating in horsepower tells you how many joules the machine can deliver every second.

A pump motor rated at 2 hp2\ \mathrm{hp} runs at its full rating for one minute. How much work does it do?

  1. A89.5 kJ89.5\ \mathrm{kJ}
  2. B1492 J1492\ \mathrm{J}
  3. C44.8 kJ44.8\ \mathrm{kJ}
  4. D120 J120\ \mathrm{J}
Show answer and solution

Answer: Option A

First the power in watts: 2×746=1492 W2 \times 746 = 1492\ \mathrm{W}, which is 14921492 joules every second. Over 60 s60\ \mathrm{s} that is W=Pt=1492×60=89520 J≈89.5 kJW = Pt = 1492 \times 60 = 89520\ \mathrm{J} \approx 89.5\ \mathrm{kJ}. Option B stops at the power and calls it work — a watt is not a joule until a time multiplies it. C uses one horsepower instead of two, and D multiplies 2×602 \times 60 as if a horsepower were a watt.

Q5One correct option

A lift motor is the car problem turned vertical. At steady speed the cage does not accelerate, so the cable must pull with exactly the forces pulling the other way, and the motor's power is that pull times the speed.

A lift cage and its passengers together have mass 500 kg500\ \mathrm{kg}. Friction is negligible. What power must the motor deliver to raise it at a steady 2 m/s2\ \mathrm{m/s}? Take g=10 m/s2g = 10\ \mathrm{m/s^{2}}.

  1. A1 kW1\ \mathrm{kW}
  2. B5 kW5\ \mathrm{kW}
  3. C10 kW10\ \mathrm{kW}
  4. D20 kW20\ \mathrm{kW}
Show answer and solution

Answer: Option C

At steady speed the tension balances the weight: T=Mg=500×10=5000 NT = Mg = 500 \times 10 = 5000\ \mathrm{N}. The cable moves up at 2 m/s2\ \mathrm{m/s} in the direction it pulls, so P=Tv=5000×2=10000 W=10 kWP = Tv = 5000 \times 2 = 10000\ \mathrm{W} = 10\ \mathrm{kW}. Option B is the tension in kilonewtons, a force mistaken for a power. A drops gg and uses the mass as if it were the weight, and D doubles the right answer, as if the speed were counted twice. Another way to see it: each second the cage rises 2 m2\ \mathrm{m}, so the motor does mgh=500×10×2mgh = 500 \times 10 \times 2 joules every second — the same 10 kW10\ \mathrm{kW} from the first rung's idea.

Q6Numerical answer

Now the other way round: hold the power fixed and see what the motion does. A body starting from rest on a smooth level floor, driven by a source of constant power PP, receives PtPt joules in a time tt — and with no other force doing work on it, all of that becomes kinetic energy: 12mv2=Pt\tfrac{1}{2}mv^{2} = Pt.

A 2 kg2\ \mathrm{kg} body starts from rest on a smooth level floor and is driven at a constant 16 W16\ \mathrm{W}. What is its speed after 4 s4\ \mathrm{s}, in m/s\mathrm{m/s}?

Show answer and solution

Answer: 8 m/s

In 4 s4\ \mathrm{s} the source delivers 16×4=64 J16 \times 4 = 64\ \mathrm{J}, all of it kinetic energy: 12×2×v2=64\tfrac{1}{2} \times 2 \times v^{2} = 64, so v2=64v^{2} = 64 and v=8 m/sv = 8\ \mathrm{m/s}. Notice what was not used: no force, no acceleration, no equation of motion — those assume a constant acceleration, and at constant power the acceleration is anything but constant. Writing v=atv = at with some fixed aa is the trap this whole rung is built against.