JEE Main chapters · Class 11 · Physics

Work, Energy and Power: JEE Main Physics formulas and notes

Work, Energy and Power is the chapter where mechanics comes together: Laws of Motion, Rotation, Gravitation and Oscillations all use it. JEE Mains usually gives 2 to 3 questions from it. The favourites are the work-energy theorem with a variable force, energy conservation with a spring, the vertical circle and collisions with the coefficient of restitution. Aim for 90 seconds per question. About 10 hours in total is enough.

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Open Physica, choose Rank mode, JEE Mains, then Physics and this chapter for the guide, formulas, most-asked questions, past-year questions and a mock test.

Quick revision

• W = Fs cosθ, W = ∫F dx = area under the F–x graph
• W(net) = ΔK = ½m(v² − u²)
• K = (p²)/(2m), same momentum: K ∝ (1)/(m)
• U(spring) = ½kx², F = −(dU)/(dx)
• Conservation: K₁ + U₁ = K₂ + U₂ (heat = μN × path)
• Block dropped on a spring: mg(h + x) = ½kx²
• P = (W)/(t) = F⃗·v⃗, P = Fv, top speed v = (P)/(R)
• Jet: P = ½ρAv³
• Vertical circle: bottom √(5gr), top √(gr), T(bottom) − T(top) = 6mg
• Loop from rest: h = 2.5r; rod: bottom 2√(gr)
• Hemisphere: leaves at cosθ = (2)/(3)
• Elastic: equal masses swap velocities, e = (v₂ − v₁)/(u₁ − u₂)
• Inelastic: v = (m₁u₁ + m₂u₂)/(m₁ + m₂), loss = (m₁m₂)/(2(m₁ + m₂))(u₁ − u₂)²(1 − e²)
• Bounce: e = √(h₂)/(h₁), h(n) = e²ⁿh₁

Formulas with solved examples

Work

Work done by a constant force

• W = F⃗·s⃗ = Fs cosθ
• work is a scalar: positive for θ < 90°, zero at 90°, negative for θ > 90°
• work done by gravity = ±mgh and does not depend on the path
• net work = sum of the work of all forces

How to apply. Find the angle between the force and the displacement, not between the force and the surface. Friction and drag usually do negative work. A force at right angles to the motion (the normal force, the tension in a circle) does no work.

Example. A 25 kg block is pulled at constant speed over 5 m by a force at 45° to the horizontal, μ = 0.25 (g = 9.8): F = 49√2 N and W = F cos45° × 5 = 245 J.

Work done by a variable force

• W = ∫F dx
• work = area under the F–x graph
• F = α + βx²: W = αx + (βx³)/(3)
• area below the axis is negative work
• if F depends on position, use the integral, not Fs

How to apply. Write F as a function of x, integrate between the limits, or add the areas of triangles, rectangles and trapeziums on the graph. For F that rises uniformly from zero, the average force is half the final force (in general it is the mean of the first and last values).

Example. F = α + βx² with α = 1 N: W = 5 J for 1 m gives 1 + β/3 = 5, so β = 12 N/m².

Work against friction and gravity

• on a level surface: W(friction) = −μmg s
• on an incline: W(friction) = −μmg cosθ × s
• W(gravity) = −mg sinθ × s going up
• work by gravity depends only on the height, W = −mgh going up

How to apply. Friction work depends on the length of the path, gravity work only on the height. Work done against a force is the negative of the work done by it.

Example. 100 kg block slid 10 m on a floor with μ = 0.4 (g = 10): work against friction = 0.4 × 1000 × 10 = 4000 J.

Work done by a spring

• W(spring) = −½k(x₂² − x₁²)
• work you do to stretch it from 0 to x = ½kx²
• the F–x graph of a spring is a straight line through the origin, so the work is a triangle

How to apply. Use the extension from the natural length, not the length of the spring. The spring does negative work while it is being stretched, and positive work when it returns.

Example. A spring of k = 200 N/m stretched from 0.1 m to 0.3 m: work done by you = ½ × 200 × (0.09 − 0.01) = 8 J.

Lifting a chain, rope or a pile

• work to lift a hanging part = (its mass) × g × (rise of its centre of mass)
• chain with length l hanging: W = (Mg l²)/(2L)
• 1/n of the cable hanging: W = (MgL)/(2n²)
• n bricks of height h piled up: W = ½ n(n − 1) mgh

How to apply. The centre of mass of the hanging part is half its length below the edge. Raise only that part, and the rise of its centre of mass is that distance.

Example. A 4 m rope of mass 8 kg with 1 m hanging (g = 10): W = 8 × 10 × 4 ÷ (2 × 16) = 10 J.

Kinetic energy

Kinetic energy and the work-energy theorem

• K = ½mv² = (p²)/(2m)
• p = √(2mK)
• W(net) = ΔK = ½m(v² − u²)
• the theorem holds for any force, constant or variable, conservative or not

How to apply. Add the work of all the forces, including friction and gravity, then equate it to the change in kinetic energy. Energy and work are scalars, so there is no direction to worry about.

Example. A 0.5 kg body has v = 3x² + 4 m/s: from x = 0 to 2 m, v goes from 4 to 16 m/s and W = ½ × 0.5 × (256 − 16) = 60 J.

Kinetic energy and momentum

• same momentum: K ∝ (1)/(m)
• same kinetic energy: p ∝ √m
• doubling the speed makes K four times
• p increased by x%: K increases by about 2x% when x is small
• kinetic energy in the centre of mass frame = ½μ v(rel)², with μ = (m₁m₂)/(m₁ + m₂)

How to apply. Eliminate the speed: K = p²/2m. Two bodies with the same momentum have kinetic energy inversely proportional to their masses.

Example. Masses 1 kg and 4 kg with the same momentum: K₁ : K₂ = 4 : 1. If p rises by 1%, K rises by about 2%.

Stopping distance and penetration

• work against a constant resistance: F d = ½mv²
• d ∝ v²
• for identical planks, each takes away the same kinetic energy: ΔK = F d per plank, so v² falls by equal steps
• work done by friction on a stopping vehicle = its initial kinetic energy

How to apply. Set the work done by the resisting force equal to the kinetic energy lost. Each plank of the same thickness takes away the same kinetic energy F d, so v² falls by the same amount per plank (it is not the same fraction of speed).

Example. A 20 g bullet at 100 m/s stops in 0.1 m: F = ½ × 0.02 × 10⁴ ÷ 0.1 = 1000 N.

Potential energy

Potential energy and the force from it

• gravity near the ground: U = mgh
• spring: U = ½kx²
• F = −(dU)/(dx)
• equilibrium where dU/dx = 0: stable if U is a minimum, unstable if a maximum
• for a conservative force, work around a closed path is zero

How to apply. The force is the negative slope of the U–x graph. A body is pushed towards lower potential energy. Only a difference of potential energy matters, so the zero level can be chosen freely.

Example. U = 3x² − 6x: F = −(6x − 6) and F = 0 at x = 1 m, where U is a minimum, so it is stable.

Conservative and non-conservative forces

• conservative: gravity, spring force, electrostatic force
• non-conservative: friction, air drag, viscous force
• W(conservative) = −ΔU
• W(friction) = −μN × (length of the path)

How to apply. A force is conservative if the work does not depend on the path. Friction depends on the path, so it turns mechanical energy into heat.

Example. Moving a block 10 m across a floor and back with f = 5 N: the work of friction is −100 J, not zero.

Springs in series and in parallel

• parallel: k = k₁ + k₂
• series: (1)/(k) = (1)/(k₁) + (1)/(k₂)
• a spring cut into n equal parts: each part has nk
• same stretch: U ∝ k
• same force: U ∝ (1)/(k)

How to apply. In parallel both springs stretch equally. In series both carry the same force and the stretches add up.

Example. Two springs of 200 N/m in series give k = 100 N/m. Stretched by 0.2 m the combination stores ½ × 100 × 0.04 = 2 J.

Conservation of energy

Conservation of mechanical energy

• K₁ + U₁ = K₂ + U₂ when only conservative forces do work
• fall through a height h: v = √(2gh)
• with friction: K₁ + U₁ = K₂ + U₂ + heat
• heat produced = μN × (path length)

How to apply. Choose a reference level for U and write the total energy at the two points. Forces that do no work (the normal force, the tension of a string in a circle) can be ignored.

Example. A ball dropped from 10 m (g = 9.8) hits the floor at √(2 × 9.8 × 10) = 14 m/s. If it rebounds to 5 m, it keeps 50% of its energy.

Spring and block problems

• dropped on a spring from a height h: mg(h + x) = ½kx²
• x(max) = (mg + √(m²g² + 2mghk))/(k)
• a block released on a spring at its natural length: x(max) = (2mg)/(k), equilibrium at (mg)/(k)
• speed at the natural length after compression x: v = x√(k)/(m)

How to apply. At the lowest point the speed is zero, so all the lost gravitational energy is in the spring. The equilibrium position is where the speed is largest, not where it stops.

Example. A 2 kg block is released on a spring (k = 200 N/m) at its natural length: x(max) = 2 × 2 × 10 ÷ 200 = 0.2 m and the equilibrium is at 0.1 m (g = 10).

Slides, slopes and rough floors

• smooth slide of height h: v = √(2gh)
• rough incline: ½mv² = mgh − μmg cosθ × s
• stopping distance on a rough floor: s = (v²)/(2μg)
• a body released from height h down a smooth curve that stops on a rough horizontal floor of coefficient μ: distance on the floor = (h)/(μ)

How to apply. Write the energy equation from the start to the end, putting the work of friction on the right. On a floor that is flat the whole loss of energy equals μmg × path.

Example. A block slides from rest down a smooth incline of height 5 m, then across a rough floor of μ = 0.5: v² = 100, so it stops after 100 ÷ (2 × 0.5 × 10) = 10 m (g = 10).

Chain sliding off a smooth table

• a chain of length L with length l hanging, released: v = √(g(L² − l²))/(L)
• use the centre of mass of each part for U
• potential energy of the hanging part = −(mgl²)/(2L)

How to apply. Take the table top as the zero of potential energy. The gain in kinetic energy equals the loss of potential energy of the whole chain.

Example. L = 1 m, l = 0.5 m (g = 10): v = √(10 × 0.75) = 2.74 m/s when it just leaves the table.

Power

Power

• P = (W)/(t) = (dW)/(dt) = F⃗·v⃗
• average power = total work ÷ total time
• 1 watt = 1 J/s, 1 hp = 746 W
• 1 kWh = 3.6 × 10⁶ J

How to apply. Instantaneous power is force times velocity at that moment. Average power is the whole work over the whole time, which is not the average of the power values unless they change uniformly.

Example. A pump lifts 1000 kg of water per minute through 6 m (g = 10): P = 1000 × 10 × 6 ÷ 60 = 1000 W.

Power of an engine and efficiency

• P = Fv
• at constant speed against a resistance R: P = Rv
• constant power, no resistance: a = (P)/(mv)
• with resistance R: a = (P/v − R)/(m)
• efficiency η = (useful output)/(input)
• power rises as t for constant acceleration

How to apply. The driving force falls as the speed rises at constant power, so the acceleration falls until the force equals the resistance. Then the speed is steady at v = P/R.

Example. A 20 kW engine against a resistance of 500 N: top speed = 20000 ÷ 500 = 40 m/s.

Variable power and constant acceleration

• W = ∫P dt
• F = kt from rest: v = (kt²)/(2m) and P = (k²t³)/(2m)
• constant acceleration a from rest: P = m a²t
• work in time t = ½m a²t²

How to apply. Write v(t) from the motion, then P = Fv. The work in a time interval is the area under the P–t graph.

Example. A 2 kg body accelerates uniformly at 3 m/s² from rest: at t = 2 s, F = 6 N, v = 6 m/s, so P = 36 W. Work in 2 s = ½ × 2 × 6² = 36 J.

Pumps, lifts and flowing water

• water lifted: P = (m)/(t) g h
• a jet or a flow: P = ½ρAv³
• wind turbine: P = ½ρAv³η
• kinetic energy per second of a flow = ½(mass per second)v²

How to apply. The mass flowing each second is ρAv. Multiply by ½v² for the kinetic energy per second, or by gh for the potential energy per second.

Example. A jet of water with A = 10⁻³ m² and v = 5 m/s: P = ½ × 1000 × 10⁻³ × 125 = 62.5 W.

Constant force on a body from rest

• v = (Ft)/(m)
• K = (F²t²)/(2m)
• P = Fv = (F²t)/(m)
• average power over t = (F²t)/(2m)

How to apply. Power rises in proportion to t, and the average power is half the final power.

Example. F = 6 N on m = 2 kg for 2 s: v = 6 m/s, P = 36 W at t = 2 s, average power = 18 W, work = 36 J.

Vertical circle

Vertical circle with a string

• at the bottom: T − mg = (mv²)/(r)
• at the top: T + mg = (mv²)/(r)
• to complete the circle: v(top) ≥ √(gr) and v(bottom) ≥ √(5gr)
• T(bottom) − T(top) = 6mg
• v² = v₀² − 2gr(1 − cosθ) at angle θ from the bottom

How to apply. The string goes slack when the tension at the top would be zero. Use energy conservation between the bottom and the top, and the force equation at each point.

Example. r = 0.9 m (g = 10): minimum speed at the bottom = √45 = 6.7 m/s and at the top 3 m/s. The tension at the bottom is 6mg more than at the top.

Loop, bead and rod

• rod or tube: v(top) ≥ 0 and v(bottom) ≥ 2√(gr)
• track (cart or ball) loop: leaves the track where the normal force N = 0
• a block or ball sliding (not rolling) from rest on a smooth track must start at h = 2.5r above the bottom of a loop of radius r
• speed at the top of a loop from height h: v² = 2g(h − 2r)

How to apply. A rod can push as well as pull, so the top speed can be zero. A track can only push, so the condition N ≥ 0 at the top gives v ≥ √(gr).

Example. A block sliding down a smooth track needs h = 2.5 r to complete a loop of radius r = 0.4 m, which is 1 m.

Block on a smooth hemisphere

• a block released from the top leaves the surface where cosθ = (2)/(3) (θ from the vertical)
• at that point v² = (2gR)/(3)
• it leaves at a height (2R)/(3) above the base
• N = mg(3cosθ − 2) before it leaves

How to apply. Write mg cosθ − N = mv²/R and the energy equation v² = 2gR(1 − cosθ), then put N = 0.

Example. A block slides from the top of a smooth hemisphere of radius 1.5 m: it leaves at a height of 1 m above the base, with v = √(2 × 10 × 1.5 ÷ 3) = √10 m/s.

Pendulum and the slack string

• T = (mv²)/(r) + mg cosθ at angle θ from the lowest point
• bottom speed v ≤ √(2gr): the body oscillates and never rises above the horizontal
• √(2gr) < v < √(5gr): the string goes slack between the horizontal and the top
• v(bottom) = √(2gr) just reaches the horizontal

How to apply. Compare the bottom speed with √(2gr) and √(5gr) first. The string can go slack only when the body is above the centre.

Example. r = 1 m, v at the bottom = 4 m/s (g = 10): 4 < √20 = 4.47, so the body stays below the horizontal and rises to h = v²/2g = 0.8 m.

Collisions

Collisions and the coefficient of restitution

• momentum is always conserved: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
• e = (v₂ − v₁)/(u₁ − u₂)
• elastic: e = 1 and kinetic energy is conserved
• perfectly inelastic: e = 0 and the bodies stick: v = (m₁u₁ + m₂u₂)/(m₁ + m₂)

How to apply. Write momentum conservation first. Use e to get the second equation. Take one direction as positive and give the velocities their signs.

Example. m₁ = 3 kg at 4 m/s and m₂ = 1 kg at rest stick together: v = 12 ÷ 4 = 3 m/s.

Elastic collision in one dimension

• v₁ = ((m₁ − m₂)u₁ + 2m₂u₂)/(m₁ + m₂)
• v₂ = ((m₂ − m₁)u₂ + 2m₁u₁)/(m₁ + m₂)
• equal masses exchange velocities
• a light body hitting a much heavier body at rest bounces back at almost the same speed (exactly, for a body of infinite mass)
• fraction of energy transferred to a body at rest = (4m₁m₂)/((m₁ + m₂)²)

How to apply. Learn the special cases: equal masses, a heavy body hitting a light one at rest (the light one goes at 2u) and the reverse. Energy passed on is greatest for equal masses.

Example. m₁ = 2 kg at 6 m/s hits m₂ = 4 kg at rest: v₁ = (2 − 4) × 6 ÷ 6 = −2 m/s and v₂ = 2 × 2 × 6 ÷ 6 = 4 m/s.

Inelastic collision and the loss of energy

• perfectly inelastic: loss of K = (m₁m₂)/(2(m₁ + m₂))(u₁ − u₂)²
• general: loss = (m₁m₂)/(2(m₁ + m₂))(u₁ − u₂)²(1 − e²)
• bullet stuck in a block: fraction of K lost = (M)/(m + M)

How to apply. The common velocity comes from momentum conservation. The loss is the difference of kinetic energy before and after, which is a fraction of the energy in the centre of mass frame.

Example. A 10 g bullet at 400 m/s embeds in a 990 g block: v = 4 m/s and 99% of the kinetic energy is lost.

Bouncing ball and a wall

• ball on a floor: e = √(h₂)/(h₁)
• rebound speed = e × speed of impact
• after n bounces: h(n) = e²ⁿ h₁
• a ball hitting a wall of infinite mass: v = e u

How to apply. The wall or floor has infinite mass, so only the ball moves after impact. The speed after the bounce is e times the speed before, and the heights are in a geometric ratio e².

Example. Dropped from 10 m and rebounding to 2.5 m: e = √0.25 = 0.5. After the second bounce it rises to 10 × 0.5⁴ = 0.625 m.

Collisions in two dimensions and with springs

• momentum is conserved along each axis
• along the line of impact use e
• equal masses in an oblique elastic collision, one at rest: the two paths are at 90° after the collision (in a head-on collision the first body stops)
• at maximum compression of a spring both bodies move with the same velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂) (= (m₁u₁)/(m₁ + m₂) if m₂ is at rest), and U = the loss of K

How to apply. Resolve along the line of impact and across it. Across the line of impact there is no force, so that component of each velocity stays the same. At the instant of greatest compression the relative velocity is zero.

Example. A block of mass m with speed u compresses a spring attached to a block of equal mass at rest: at maximum compression both move at u/2 and U = ½mu² − ½(2m)(u/2)² = ¼mu².

General collision with e (one dimension)

• v₁ = ((m₁ − e m₂)u₁ + m₂(1 + e)u₂)/(m₁ + m₂)
• v₂ = ((m₂ − e m₁)u₂ + m₁(1 + e)u₁)/(m₁ + m₂)
• loss of K = (m₁m₂)/(2(m₁ + m₂))(1 − e²)(u₁ − u₂)²

How to apply. Use these when e is given and neither 0 nor 1. Check e = 1 and e = 0 give the elastic and sticking results.

Example. m₁ = m₂ = 1 kg, u₁ = 4 m/s, u₂ = 0, e = 0.5: v₁ = (1 − 0.5) × 4 ÷ 2 = 1 m/s and v₂ = 1.5 × 4 ÷ 2 = 3 m/s.

Ball bouncing repeatedly (distance and time)

• total distance = h₁(1 + e²)/(1 − e²)
• total time = √(2h₁)/(g) (1 + e)/(1 − e)
• speed after n bounces = eⁿ u

How to apply. Add the geometric series of the rise and fall distances and of the flight times.

Example. Dropped from 10 m with e = 0.5 (g = 10): total distance = 10 × 1.25 ÷ 0.75 = 16.67 m and total time = √2 × 3 = 4.24 s.

Ballistic pendulum and the bullet–block rise

• bullet m at speed u stuck in a block M hanging from a string: V = (mu)/(m + M)
• rise h = (V²)/(2g)
• u = (m + M)/(m)√(2gh)

How to apply. Use momentum conservation in the short impact, then energy conservation for the swing. Energy is not conserved in the impact.

Example. m = 10 g, M = 990 g, u = 400 m/s (g = 10): V = 4 m/s and h = 16 ÷ 20 = 0.8 m.

Most asked question types

How do I solve “Work done by a constant force” questions in JEE Main?

A block is pulled, pushed or lifted. Find the work of one force or of all the forces. 1. Find the angle between the force and the displacement. 2. Use W = Fs cosθ for each force. 3. Add them for the net work, and check the signs. Example: 25 kg block pulled at 45° over 5 m at constant speed with μ = 0.25: W = 245 J.

How do I solve “Variable force and the F–x graph” questions in JEE Main?

The force depends on x or is given as a graph. Find the work or a constant in the force. 1. Write the force as a function of x, or read the graph. 2. Integrate F dx or add the areas. 3. Count the area below the axis as negative. Example: F = α + βx², α = 1 N, W = 5 J for 1 m: β = 12 N/m².

How do I solve “Work in lifting a chain, rope or pile” questions in JEE Main?

A part of a chain hangs over the edge, or books are stacked. Find the work to lift or stack them. 1. Find the mass of the part that is lifted. 2. Find the rise of its centre of mass. 3. Multiply: W = mg × rise. Example: A 4 m rope of 8 kg with 1 m hanging: W = 8 × 10 × 4 ÷ 32 = 10 J.

How do I solve “Work-energy theorem” questions in JEE Main?

A body speeds up or slows down under several forces. Find the speed, the distance or the work. 1. Write the work of each force with its sign. 2. Equate the net work to ½m(v² − u²). 3. Solve for the unknown. Example: 0.5 kg, v changes from 4 to 16 m/s: W = ½ × 0.5 × (256 − 16) = 60 J.

How do I solve “Kinetic energy and momentum” questions in JEE Main?

Two bodies have equal momentum or equal energy, or the momentum changes by a percentage. 1. Write K = p²/2m. 2. Compare using the quantity that is equal. 3. For small changes, ΔK/K ≈ 2Δp/p. Example: Equal momenta, masses 1 kg and 4 kg: K₁ : K₂ = 4 : 1.

How do I solve “Springs and elastic energy” questions in JEE Main?

A spring is stretched or compressed, or two springs are joined. Find the energy, the force or the work. 1. Use U = ½kx² with x from the natural length. 2. Combine k as k₁ + k₂ in parallel, 1/k as a sum in series. 3. The work done is the change in energy. Example: 200 N/m stretched from 0.1 m to 0.3 m: W = ½ × 200 × (0.09 − 0.01) = 8 J.

How do I solve “Conservation of energy: falls and slides” questions in JEE Main?

A body falls, slides down a smooth curve or hangs on a spring. Find the speed or the height. 1. Choose a reference level for potential energy. 2. Write K + U at the start and at the end. 3. Equate them (add the heat if there is friction). Example: A ball dropped from 10 m reaches the floor at √(2 × 9.8 × 10) = 14 m/s.

How do I solve “Friction and the loss of energy” questions in JEE Main?

A body slides on a rough floor or incline. Find the distance, the speed or the heat produced. 1. Find the work of friction: μN × path. 2. Put it in the energy equation. 3. On a slope use N = mg cosθ. Example: After a smooth slide of 5 m height, a block on a rough floor μ = 0.5 stops after 10 m.

How do I solve “Power and efficiency” questions in JEE Main?

A pump, a lift or a person does work in a given time. Find the power or the time. 1. Find the total work or energy. 2. Divide by the time: P = W/t. 3. For efficiency, divide the useful output by the input. Example: 1000 kg of water lifted 6 m per minute: P = 1000 × 10 × 6 ÷ 60 = 1000 W.

How do I solve “Vehicles and variable power” questions in JEE Main?

A vehicle runs at constant power, or a body accelerates under a changing force. Find the top speed or the power at a time. 1. Use P = Fv. 2. At top speed F equals the resistance. 3. For a changing force, find v(t) first. Example: 20 kW against 500 N: top speed 40 m/s.

How do I solve “Vertical circle and loops” questions in JEE Main?

A body on a string, a rod or a track moves in a vertical circle. Find the speed or the tension at a point, or the release height. 1. Write the force equation at the point (mv²/r). 2. Use energy conservation between the points. 3. Apply the condition: top √(gr), bottom √(5gr), loop h = 2.5r. Example: r = 0.9 m: minimum speed at the bottom √45 ≈ 6.7 m/s.

How do I solve “Elastic collision” questions in JEE Main?

Two bodies collide with no loss of kinetic energy. Find the velocities after the collision. 1. Conserve momentum. 2. Use e = 1 (or the formulas). 3. Check the special cases, such as equal masses. Example: 2 kg at 6 m/s hits 4 kg at rest: v₁ = −2 m/s, v₂ = 4 m/s.

How do I solve “Inelastic collision and energy loss” questions in JEE Main?

Bodies stick together or lose part of their energy. Find the common speed or the energy lost. 1. Use momentum conservation for the common velocity. 2. Subtract the final kinetic energy from the initial one. 3. For bullet and block, the fraction lost is M/(m + M). Example: 10 g bullet at 400 m/s in a 990 g block: v = 4 m/s, 99% of the energy is lost.

How do I solve “Bouncing ball and the coefficient of restitution” questions in JEE Main?

A ball rebounds from the floor or a wall. Find e, the rebound height or the number of bounces. 1. Write e = rebound speed ÷ impact speed. 2. Heights are in the ratio e². 3. Use h(n) = e²ⁿ h₁ for the nth rebound. Example: Dropped from 10 m, rebounding to 2.5 m: e = 0.5.

Common myths and the truth

• Myth: Work is a vector because force and displacement are vectors. Truth: Work is the dot product of two vectors, which is a scalar.
• Myth: A force that acts for a long time must do a lot of work. Truth: Work needs displacement along the force. A force at right angles to the motion does none.
• Myth: Mechanical energy is always conserved. Truth: It is conserved only if conservative forces alone do work. Friction turns it into heat.
• Myth: Kinetic energy is conserved in all collisions. Truth: Only in elastic collisions. In every collision, momentum is conserved.
• Myth: In a perfectly inelastic collision all the kinetic energy is lost. Truth: The bodies move together, so the kinetic energy of the centre of mass remains. It is the kinetic energy in the centre of mass frame that is lost (this is the maximum possible loss). If the colliding bodies move in opposite directions with equal momentum, all the kinetic energy is lost.
• Myth: The tension in a vertical circle is the same at the top and the bottom. Truth: It is larger at the bottom, by 6mg for a full circle.
• Myth: A body at the lowest point of a swing has its largest potential energy. Truth: The potential energy is least at the lowest point, and the kinetic energy is greatest.
• Myth: Power is the work done. Truth: Power is the rate of doing work. The same work in less time needs more power.
• Myth: A spring stores energy proportional to its extension. Truth: It stores ½kx², proportional to the square of the extension.
• Myth: The work done by gravity depends on the path. Truth: Gravity is conservative: its work depends only on the change in height.

Mistakes to avoid

• Work is a scalar, but its sign matters: friction does negative work.
• The angle in W = Fs cosθ is between the force and the displacement.
• The normal force and the tension of a circular path do no work.
• Mechanical energy is conserved only when no friction or other loss acts.
• Momentum is conserved in every collision, kinetic energy only in an elastic one.
• A spring stretched by x from its natural length stores ½kx², not ½k(length)².
• In a vertical circle the tension is zero at the top only for the minimum speed √(gr).
• The work done in lifting is positive for you and negative for gravity.

Simulations for Work, Energy and Power

• Ballistic pendulum — Embed a projectile in a suspended block and infer the height it can swing.
• Elastic collision — Follow two masses before and after a head-on elastic collision.
• Conservation of energy — Watch gravitational potential energy turn into kinetic energy on a frictionless track.
• Power of a lifting motor — Raise a load at constant speed and compare work with the rate of doing work.
• Glancing collision of two balls — Strike a stationary ball off-centre. Vary the impact offset and the coefficient of restitution.
• Spring launcher and energy transfer — Release a compressed spring and watch elastic potential energy turn into the cart’s kinetic energy.
• Work done by a variable force — A force that grows with displacement pushes a block. The work is the area under the F–x graph.
• Motion in a vertical circle — Whirl a stone on a string in a vertical circle. Below √(5gR) at the bottom, the string goes slack.
• Work done by a force at an angle — Change the direction of a force and see how much work it does.

More: Class 11 Physics simulations · all JEE Main chapters.