JEE Mains Physics · Work, Energy and Power

Power

Power is the rate of doing work: P = F·v at an instant, and work divided by time on average.

All Work, Energy and Power notes JEE Mains Physics strategy

Why this matters

Fourteen PYQs, nine of them multiple choice, and one from 2026. Nine use P = F·v or P = W/t: four time-dependent forces acting from rest, one average against instantaneous power, motors and a dam moving water, an elevator, and a block pulled up an incline; five drive a body with constant power and ask how its speed or position grows with time. The time-dependent force is the one to practise: velocity first, by integrating, and only then the dot product.

Concept 1 of 2: Power from force and velocity

Power is how fast work is being done. A force F on a body moving at velocity v does work at the rate F·v. If the force changes with time and the body starts from rest, the velocity is not given, so find it first by integrating the acceleration F/m.

Definition

Instantaneous: $P = \vec F\cdot\vec v$ . Average: $P_{\text{avg}} = W/t = \Delta K/t$ .
Force $\vec F(t)$ from rest: $\vec v = \dfrac{1}{m}\displaystyle\int_{0}^{t}\vec F\,dt$ , then $P = \vec F\cdot\vec v$ at the same instant.
Lifting a mass m through h in time t: $P = mgh/t$ . Water pumped or falling: $P = (\text{mass per second})\,gh$ ; a rate per hour is divided by 3600.
Efficiency η: useful power $= \eta \times$ input power.
Elevator rising at steady speed v against friction f: $P = (Mg + f)v$ .
Pulled up a smooth incline with acceleration a: $F = ma + mg\sin\theta$ , then $P = Fv$ .
Units: 1 W = 1 J/s; 1 hp = 746 W; 1 kWh = $3.6 \times 10^{6}$ J.

Power

P = \vec F\cdot\vec v \qquad P_{\text{avg}} = \frac{W}{t} \qquad \vec v = \frac{1}{m}\int_{0}^{t}\vec F\,dt

Worked example

A 3 kg body starts from rest under a force

\vec F = (6t\,\hat i + 3t^{2}\,\hat j)

N. Find the power delivered by the force at

t = 2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q24Moderate

Example 1 · Work, Energy and Power · Power

A body of mass 2 kg begins to move under the influence of time dependent force

\overrightarrow{F} =

\left( 2t\widehat{i} + 6t^{2}\widehat{j} \right)N

, where

\widehat{i}

and

\widehat{j}

are unit vectors along

x

and y -axis respectively. The power produced by the force at

t = 2\text{ }s

\_\_\_\_

Instantaneous power needs the velocity at that instant

P = F·v uses the force and the velocity at the same moment. With a time-dependent force, multiplying by an average velocity, or by v = (F/m)t as if the force were constant, gives the wrong power.

Count friction and efficiency

An elevator motor at steady speed supplies (Mg + f)v, not Mgv. A pump of efficiency η draws its useful power divided by η, which is more than the useful power, not less.

Per hour means per 3600 seconds

Water falling at 3600 kg per hour is 1 kg per second. Leaving the rate per hour in P = (mass per second)gh gives a power 3600 times too large.

Drill 8 more on power from force and velocity

Concept 2 of 2: Motion under constant power

With constant power the kinetic energy grows at a steady rate, so ½mv² = Pt. The speed then grows as the square root of time, and the distance as time to the power three halves. The force is not constant: it falls as the body speeds up, because F = P/v.

Definition

From rest: $\tfrac12 mv^{2} = Pt$ , so $v = \sqrt{\dfrac{2P}{m}}\,t^{1/2}$ .
Integrating: $x = \tfrac{2}{3}\sqrt{\dfrac{2P}{m}}\,t^{3/2} = \sqrt{\dfrac{8P}{9m}}\,t^{3/2}$ .
$a = dv/dt \propto t^{-1/2}$ and $F = P/v \propto t^{-1/2}$ .
Distances in times $t_1$ and $t_2$ from rest: $x_1 : x_2 = (t_1/t_2)^{3/2}$ .
Also $x = \tfrac{2}{3}vt$ : the distance is two thirds of what the final speed would cover in the same time.

Constant power from rest

v = \sqrt{\frac{2P}{m}}\,t^{1/2} \qquad x = \sqrt{\frac{8P}{9m}}\,t^{3/2}

Worked example

A 4 kg cart starts from rest, driven by a motor that delivers a constant 8 W. Find its speed and the distance it has covered after 9 s.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q117Moderate

Example 2 · Work, Energy and Power · Power

A body of mass

2\text{ }kg

is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 s is

\frac{1}{3}\alpha^{2}\sqrt{P}m

. The value of

\alpha

will be___

Constant power is not constant force

With constant power the force falls as the speed rises. Using v = at and x = ½at², as for a constant force, gives x ∝ t² instead of x ∝ t^(3/2).

The power is t^(3/2), not t^(2/3)

Both appear in the options. The distance grows faster than t, because the body keeps speeding up, so the power of t must be more than 1: three halves.

Drill 4 more on motion under constant power

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Power from force and velocity
Power
$P = \vec F\cdot\vec v \qquad P_{\text{avg}} = \frac{W}{t} \qquad \vec v = \frac{1}{m}\int_{0}^{t}\vec F\,dt$
Motion under constant power
Constant power from rest
$v = \sqrt{\frac{2P}{m}}\,t^{1/2} \qquad x = \sqrt{\frac{8P}{9m}}\,t^{3/2}$

Watch out for (5)

Instantaneous power needs the velocity at that instant
→ Power from force and velocity
Count friction and efficiency
→ Power from force and velocity
Per hour means per 3600 seconds
→ Power from force and velocity
Constant power is not constant force
→ Motion under constant power
The power is t^(3/2), not t^(2/3)
→ Motion under constant power

Test yourself on Work, Energy and Power

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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