PYQ Vault

JEE Mains Physics · Formula sheet

Work, Energy and Power formulas

16 formulas and 39 common traps for JEE Mains Physics Work, Energy and Power, grouped by subtopic.

Full notes with worked examples

Work Done by Constant and Variable Forces

Learn this subtopic in the notes

Work done by a constant force

Work by a constant force

W=F⃗⋅s⃗=Fscos⁡θ=Fxsx+Fysy+FzszW = \vec F\cdot\vec s = Fs\cos\theta = F_x s_x + F_y s_y + F_z s_z

Work done by a force that varies with position

Work by a variable force

W=∫x1x2F(x) dxW=∫Fx dx+∫Fy dyW = \int_{x_1}^{x_2} F(x)\,dx \qquad W = \int F_x\,dx + \int F_y\,dy

Slowing down under a retarding force that depends on position

Energy after a position-dependent retarding force

12mv2=12mu2−∫ab∣F(x)∣ dx\tfrac12 mv^{2} = \tfrac12 mu^{2} - \int_{a}^{b}|F(x)|\,dx

Common traps

A distance along a direction needs the unit vector

Moving 10 m along 3î + 4ĵ means a displacement of 6î + 8ĵ, because |3î + 4ĵ| = 5. Multiplying the direction vector by 10 gives 30î + 40ĵ, a displacement five times too long.

A force that balances friction does positive work

A box pulled at constant velocity on a rough floor has zero net work done on it. The pull still does positive work; friction does an equal negative work. Saying the applied force does zero work confuses one force's work with the net work.

Area below the axis is negative work

On an F–x graph, the parts of the curve below the x-axis are a force opposing the motion. Their area is subtracted. Adding every area as positive puts the graphs in the wrong order.

A force like x²y î depends on the path

If the x-component of a force contains y, the integral of Fₓ dx cannot be done until y is written in terms of x along the path. Integrating it with y held constant gives a number that belongs to no path.

Use the patch's end points, not its length

For F = −kx between x = 1 m and x = 3 m, the work is (k/2)(3² − 1²) = 4k. Using the length, (k/2)(2)² = 2k, treats the patch as if it began at the origin.

A retarding force takes energy away

Its work is subtracted from the starting kinetic energy. Adding it gives a final speed larger than the starting speed, which a slowing force can never produce.

Work-Energy Theorem and Kinetic Energy

Learn this subtopic in the notes

Work-energy theorem with several forces

Work-energy theorem

Wnet=∑Wi=Kf−KiW_{\text{net}} = \sum W_i = K_f - K_i

Work from a given velocity or position law

Work from two speeds

Wnet=12m(v22−v12),v=dxdtW_{\text{net}} = \tfrac12 m\left(v_2^{2} - v_1^{2}\right), \quad v = \frac{dx}{dt}

Kinetic energy and momentum, K = p²/2m

Kinetic energy and momentum

K=p22mp=2mKK = \frac{p^{2}}{2m} \qquad p = \sqrt{2mK}

Common traps

Gravity also works through the depth of penetration

A ball that falls h and then sinks d into sand is pulled down by gravity through h + d. Writing mgh = F·d leaves out mgd, and the options usually include both answers.

Constant speed does not mean every force does zero work

At constant speed the NET work is zero. The engine of a bus still does positive work, and friction does an equal negative work. The engine's work is μmgd, not zero.

Do not assume the body starts from rest

Put the starting position into the velocity law. For v = 3x² + 4 the speed at x = 0 is 4 m/s, so the starting kinetic energy is not zero. Dropping it makes the work too large.

Work done on the body, not by it

W = ΔK is the work done on the body by the forces acting on it. A question about the work done by the body on its surroundings wants the same size with the opposite sign.

Square the factor, not the percentage

Momentum up by 50% means p is multiplied by 1.5, so K is multiplied by 2.25: a rise of 125%. Doubling the 50% to get 100% is wrong.

A percentage increase subtracts the starting value

If K becomes 25 times larger, p becomes 5 times larger. That is an increase of 400%, not 500%: the new value is 500% of the old one, and the increase is 100% less.

Potential Energy and Conservation of Mechanical Energy

Learn this subtopic in the notes

Conservative forces and potential energy

Force from potential energy

F=−dUdxFx=−∂U∂xWcons=−ΔUF = -\frac{dU}{dx} \qquad F_x = -\frac{\partial U}{\partial x} \qquad W_{\text{cons}} = -\Delta U

Potential energy of a spring

Spring potential energy

U=12kx2Wx1→x2=12k(x22−x12)U = \tfrac12 kx^{2} \qquad W_{x_1 \to x_2} = \tfrac12 k\left(x_2^{2} - x_1^{2}\right)

Conservation of mechanical energy

Mechanical energy conserved

12mv12+mgh1=12mv22+mgh2vbottom2=5gL\tfrac12 mv_1^{2} + mgh_1 = \tfrac12 mv_2^{2} + mgh_2 \qquad v_{\text{bottom}}^{2} = 5gL

Common traps

Work is the integral of F·dr; the minus sign belongs to ΔU

The work done by a force from r₁ to r₂ is W = ∫F·dr. It is the change in potential energy that carries the minus sign: ΔU = −∫F·dr. A statement that writes W = −∫F·dr is false.

The force is the slope, not the height

On a U(x) graph, a steep stretch means a strong force and a flat stretch means no force, however high it sits. Ranking forces by the height of U instead of the steepness gives the wrong order.

Friction has no potential energy

The work done by friction depends on the path, so no function U can store it. Coulomb, gravitational and spring forces all have a potential energy; friction does not.

Spring energy is half of force times stretch

The spring force grows from 0 to kx as it is stretched, so the stored energy is ½kx², not kx·x. Writing kx² doubles every answer.

A dropped ball falls the extra compression too

A ball dropped from h onto a spring platform loses mg(h + x) of potential energy by the lowest point, not mgh. Using mgh gives a spring constant that is too small.

Energy between two stretches uses x₂² − x₁²

Stretching from 2 cm to 4 cm takes ½k(4² − 2²) in cm², not ½k(4 − 2)². Every extension is measured from the natural length.

A rod is not a string

A bob on a string needs a speed of √(gL) at the top, so √(5gL) at the bottom. A bob on a rigid rod can reach the top with zero speed, so √(4gL) at the bottom is enough.

Measure every height from one level

Pick the lowest point as zero and keep it. On a circle of radius R, a point at angle θ from the bottom is R(1 − cos θ) up; a point at angle θ from the top is R(1 + cos θ) up.

Both masses on a pulley carry kinetic energy

When one mass falls and the other rises, the energy released, (m₁ − m₂)gh, is shared by both: ½(m₁ + m₂)v². Giving it all to the falling mass makes the speed too large.

Power

Learn this subtopic in the notes

Power from force and velocity

Power

P=F⃗⋅v⃗Pavg=Wtv⃗=1m∫0tF⃗ dtP = \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=2Pm t1/2x=8P9m t3/2v = \sqrt{\frac{2P}{m}}\,t^{1/2} \qquad x = \sqrt{\frac{8P}{9m}}\,t^{3/2}

Common traps

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.

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.

Impulse, Explosions and Perfectly Inelastic Collisions

Learn this subtopic in the notes

Impulse and change in momentum

Impulse

J⃗=∫F⃗ dt=Δp⃗Favg=ΔpΔt\vec J = \int \vec F\,dt = \Delta\vec p \qquad F_{\text{avg}} = \frac{\Delta p}{\Delta t}

Conservation of momentum in explosions, recoil and sticking collisions

Momentum conserved

m1u⃗1+m2u⃗2=m1v⃗1+m2v⃗2ΔKlost=12m1m2m1+m2(u1−u2)2m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2 \qquad \Delta K_{\text{lost}} = \frac12\frac{m_1m_2}{m_1 + m_2}\left(u_1 - u_2\right)^{2}

Bullet and pendulum: momentum in the impact, energy in the swing

Ballistic pendulum

mu=(M+m)VV=2ghVfull circle=5gLmu = (M + m)V \qquad V = \sqrt{2gh} \qquad V_{\text{full circle}} = \sqrt{5gL}

Common traps

A rebound adds the two speeds

Momentum reverses on a rebound, so the change is m(u + v). Subtracting the speeds treats the ball as if it carried on in the same direction, and gives far too small an impulse.

At an angle, only the normal part reverses

A wall pushes only along its normal. A ball hitting at 45° to the normal has an impulse of 2mv cos 45°, which is 1/√2 of the impulse for a head-on hit at the same speed.

Same impulse, smaller force

A body brought to rest from the same speed always has the same impulse. A longer stopping time lowers the average force, not the impulse.

Momenta add as vectors

Two fragments with momentum p each, flying at right angles, have a total momentum of √2·p, not 2p. The third fragment must carry √2·p back the other way.

Kinetic energy is not conserved when bodies stick

Only momentum carries through a sticking collision. Writing ½m₁u₁² = ½(m₁ + m₂)v² gives a wrong common speed; the lost energy goes into heat, sound and deformation.

An explosion can increase kinetic energy

When a moving block splits, the pieces can together have more kinetic energy than the block had. Momentum is conserved; kinetic energy is not, in either direction.

Do not conserve energy through the impact

Setting ½mu² = (M + m)gh skips the impact and claims the bullet's kinetic energy all goes into the swing. Most of it becomes heat. Use momentum for the impact, then energy for the swing.

A bullet that bounces back gives the bob extra momentum

If the bullet recoils at v, the bob's momentum is MV = m(u + v), not m(u − v). The minus sign of the recoiling velocity turns into a plus.

Elastic Collisions and Coefficient of Restitution

Learn this subtopic in the notes

Head-on elastic collisions

Elastic collision with a target at rest

v1=m1−m2m1+m2 uv2=2m1m1+m2 uK2K1=4m1m2(m1+m2)2v_1 = \frac{m_1 - m_2}{m_1 + m_2}\,u \qquad v_2 = \frac{2m_1}{m_1 + m_2}\,u \qquad \frac{K_2}{K_1} = \frac{4m_1m_2}{(m_1 + m_2)^{2}}

Coefficient of restitution and bouncing balls

Coefficient of restitution

e=v2−v1u1−u2h′=e2hK′K=e2e = \frac{v_2 - v_1}{u_1 - u_2} \qquad h' = e^{2}h \qquad \frac{K'}{K} = e^{2}

Common traps

Exchange of velocities happens one collision at a time

With three identical balls in a line, apply the exchanges in the order the collisions happen. Swapping the first and last velocities at once skips the middle ball and gives the wrong final set.

Maximum compression is at the common velocity

Two blocks with a spring between them squeeze it most when they move at the same speed, not when one of them stops. At that instant the kinetic energy missing from the pair is stored in the spring.

A light body bounces back from a heavy one

When m₁ < m₂, v₁ = (m₁ − m₂)u/(m₁ + m₂) is negative: the light body reverses. Taking its speed as positive in the momentum equation gives a wrong starting speed.

Height goes with e², speed with e

A ball that leaves the floor at e times its arrival speed rises to e² times its starting height. Using e for the height ratio gives a rebound that is too high.

Every bounce is travelled twice

After the first fall, each bounce goes up and comes back down. The total distance is h + 2e²h + 2e⁴h + …, which sums to h(1 + e²)/(1 − e²); counting each bounce once gives too little.

More JEE Mains Physics formula sheets