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  3. Conservation of Mechanical Energy

Work, Energy and Power · JEE & NEET Physics

Conservation of Mechanical Energy: notes and previous year questions

KE + PE stays constant when only conservative forces do work: falls, slides, pendulums, springs, pulleys, and what changes when friction acts.

Conservation of Mechanical Energy in short

  • Mechanical energy is KE + PE; it stays constant when only conservative forces do work.
  • A conservative force's work is path-independent and zero around a closed loop; it has a potential energy.
  • Gravity, springs and electric forces are conservative; friction, air resistance and viscous drag are not.
  • A fall or smooth slide from height h gives v = √(2gh), whatever the path.

1Energy changes form

A roller-coaster cart that starts at rest 5 m up keeps trading energy: going down, potential energy becomes kinetic energy; going up, it turns back. On a smooth track the total never changes, so the cart can never climb above 5 m.

Mechanical energy is kinetic plus potential energy, E=KE+PEE = KE + PE.

KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_fWhen only conservative forces do work.

Example: the cart starting at rest 5 m up reaches the lowest point, 0.3 m up, with 12v2=10×4.7\tfrac{1}{2}v^2 = 10 \times 4.7, so v=94≈9.7v = \sqrt{94} \approx 9.7 m/s. The mass cancels.

2Conservative and non-conservative forces

A force is conservative if any one of these (equivalent) statements holds:

  1. Its work depends only on the start and end points, not on the path.
  2. Its work around any closed loop is zero.
  3. It has a potential energy: WA→B=UA−UBW_{A \to B} = U_A - U_B.
ConservativeNon-conservative
gravityfriction
spring forceair resistance
electric forceviscous drag in a liquid

Carry a 2 kg ball 4 m higher along a straight path or a wiggly one: gravity does −80 J either way. Friction would take more energy on the longer path. On a round trip back to the start, gravity's total work is zero.

3Falls, throws and slides

v=2ghv = \sqrt{2gh}A fall from rest through height h: mgh = ½mv².
hmax⁡=v22gh_{\max} = \frac{v^2}{2g}A throw straight up at speed v: ½mv² = mgh.

A 2 kg ball dropped from 5 m: 100 J of potential energy becomes 100 J of kinetic energy, so v=10v = 10 m/s. A ball thrown up at 20 m/s rises 400/20=20400/20 = 20 m.

Example: a ball thrown at 10 m/s (in any direction) from a 15 m roof lands with 12v2=12(10)2+10×15=200\tfrac{1}{2}v^2 = \tfrac{1}{2}(10)^2 + 10 \times 15 = 200, so v=20v = 20 m/s.

4The pendulum

A pendulum of length 2 m released from the horizontal falls 2 m to its lowest point: 12mv2=mg(2)\tfrac{1}{2}mv^2 = mg(2), so v=40≈6.32v = \sqrt{40} \approx 6.32 m/s.

The string also pulls on the bob, but always along its length, while the bob moves along the circle at 90° to the string. So the tension does no work, and only gravity changes the energy.

Released at 60° from the vertical, the bob falls L(1−cos⁡60∘)=1L(1 - \cos 60^\circ) = 1 m, so v=20≈4.47v = \sqrt{20} \approx 4.47 m/s at the bottom. The bob is fastest at the lowest point and momentarily at rest at the ends of its swing.

5Springs and heights

Springs join the sum: add 12kx2\tfrac{1}{2}kx^2 to the potential energy.

v=xkmv = x\sqrt{\frac{k}{m}}A spring launch on a smooth floor: ½kx² = ½mv².
12kx2=mgh\tfrac{1}{2}kx^2 = mghStraight up, with h measured from the lowest (squeezed) point.

6When friction acts

With friction, mechanical energy is not conserved: some of it becomes heat. Total energy is still conserved, so we keep track of the heat.

Wnc=ΔEmech=(KEf+PEf)−(KEi+PEi)W_{nc} = \Delta E_{\text{mech}} = (KE_f + PE_f) - (KE_i + PE_i)W_nc is the work of the non-conservative forces.
KEi+PEi−fs=KEf+PEfKE_i + PE_i - fs = KE_f + PE_fFor friction f over a distance s; fs is the energy turned into heat.

7Connected bodies

For bodies joined by a string, write one energy equation for the whole system. The tension's works on the two bodies cancel.

m1gh−m2gh=12(m1+m2)v2m_1gh - m_2gh = \tfrac{1}{2}(m_1 + m_2)v^2Two masses over a light, smooth pulley, from rest; m₁ falls h and m₂ rises h.

3 kg and 2 kg: 30−20=12(5)v230 - 20 = \tfrac{1}{2}(5)v^2 after 1 m, so v=2v = 2 m/s. 5 kg and 3 kg after 2 m: 40=12(8)v240 = \tfrac{1}{2}(8)v^2, so v≈3.16v \approx 3.16 m/s.

8When to use it

  • Use it for questions linking heights, stretches and speeds: falls, throws, slides, pendulums, springs, roller coasters.
  • With friction, use the modified equation.
  • Not for times or accelerations: use kinematics and Newton's laws.
  • Not across a sudden collision or explosion: kinetic energy is usually lost there, so use momentum.
  1. Choose the system and a zero level for potential energy.
  2. Sort the forces into conservative and non-conservative.
  3. Write the energy at the start and at the end, adding WncW_{nc} if any.
  4. Solve.

Summary

Key ideas

  • Mechanical energy is KE + PE; it stays constant when only conservative forces do work.
  • A conservative force's work is path-independent and zero around a closed loop; it has a potential energy.
  • Gravity, springs and electric forces are conservative; friction, air resistance and viscous drag are not.
  • A fall or smooth slide from height h gives v = √(2gh), whatever the path.
  • A throw straight up at v rises v²/2g.
  • A pendulum's string does no work, because it is perpendicular to the motion.
  • Add ½kx² for springs; heights from a spring launch are measured from the lowest point.
  • With friction, W_nc = ΔE_mech; the lost mechanical energy becomes heat.
  • For connected bodies, write one energy equation for the whole system.
  • Energy methods give heights and speeds, not times; across collisions use momentum.

Every equation

Conservation of mechanical energy
KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_f
With non-conservative forces
Wnc=ΔEmechW_{nc} = \Delta E_{\text{mech}}
With friction
KEi+PEi−fs=KEf+PEfKE_i + PE_i - fs = KE_f + PE_f
Conservative force
WA→B=UA−UBW_{A \to B} = U_A - U_B
Fall from height h
v=2ghv = \sqrt{2gh}
Throw straight up
hmax⁡=v22gh_{\max} = \frac{v^2}{2g}
Spring launch
v=xk/mv = x\sqrt{k/m}
Spring throws a block up
12kx2=mgh\tfrac{1}{2}kx^2 = mgh
Pendulum from angle θ
v=2gL(1−cos⁡θ)v = \sqrt{2gL(1 - \cos\theta)}
Rough slope
mgh−μmgcos⁡θ s=12mv2mgh - \mu mg\cos\theta\,s = \tfrac{1}{2}mv^2
Two masses on a pulley
(m1−m2)gh=12(m1+m2)v2(m_1 - m_2)gh = \tfrac{1}{2}(m_1 + m_2)v^2
Connected bodies with friction
ΔKE+ΔPE=Wfriction\Delta KE + \Delta PE = W_{\text{friction}}

Previous year questions with solutions

Real JEE and NEET questions on conservation of mechanical energy. Try each one before you open the solution.

Q1JEE Main 2025One correct option

A particle is released from height SS above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

  1. AS4,3gS2\frac{S}{4},\frac{3\mathrm{gS}}{2}
  2. BS2,3gS2\frac{S}{2},\frac{3\mathrm{gS}}{2}
  3. CS4,3gS2\frac{S}{4},\sqrt{\frac{3\mathrm{gS}}{2}}
  4. DS2,3gS2\frac{S}{2},\sqrt{\frac{3\mathrm{gS}}{2}}
Show answer and solution

Answer: Option C

Split the total mgSmgS into four equal parts: K=3UK = 3U means three of them are kinetic and one is potential. So U=mgh=14mgSU = mgh = \tfrac{1}{4}mgS and h=S4h = \dfrac{S}{4}. The particle has fallen 3S4\dfrac{3S}{4}, so 12mv2=mg×3S4\tfrac{1}{2}mv^{2} = mg \times \dfrac{3S}{4} and v=3gS2v = \sqrt{\dfrac{3gS}{2}}. Options A and B give 3gS2\dfrac{3gS}{2} without the square root — that is v2v^{2}, not a speed, and its units give it away. B and D put the particle at S/2S/2, which is where K=UK = U, not K=3UK = 3U.

Q2NEET 2025One correct option

A bob of heavy mass mm is suspended by a light string of length ll. The bob is given a horizontal velocity v0v_{0} at its lowest point. If the string gets slack at some point PP making an angle θ\theta from the horizontal, above the point of suspension, the ratio of the speed vv of the bob at point PP to its initial speed v0v_{0} is:

  1. A(cos⁡θ2+3sin⁡θ)12{(\frac{\cos \theta }{2+3\sin \theta })}^{\frac{1}{2}}
  2. B(sin⁡θ2+3sin⁡θ)12{(\frac{\sin \theta }{2+3\sin \theta })}^{\frac{1}{2}}
  3. C(sin⁡θ)12(\sin \theta {)}^{\frac{1}{2}}
  4. D(12+3sin⁡θ)12{(\frac{1}{2+3\sin \theta })}^{\frac{1}{2}}
Show answer and solution

Answer: Option B

Slack means T=0T = 0 at P, where the part of the weight pointing at the pivot is mgsin⁡θmg\sin\theta: mgsin⁡θ=mv2lmg\sin\theta = \dfrac{mv^{2}}{l}, so v2=glsin⁡θv^{2} = gl\sin\theta. P is l(1+sin⁡θ)l(1 + \sin\theta) above the starting point, so energy gives v02=v2+2gl(1+sin⁡θ)=glsin⁡θ+2gl+2glsin⁡θ=gl(2+3sin⁡θ)v_{0}^{2} = v^{2} + 2gl(1 + \sin\theta) = gl\sin\theta + 2gl + 2gl\sin\theta = gl(2 + 3\sin\theta). Dividing, v2v02=sin⁡θ2+3sin⁡θ\dfrac{v^{2}}{v_{0}^{2}} = \dfrac{\sin\theta}{2 + 3\sin\theta}, and the ratio of the speeds is its square root. Option D has lost the sin⁡θ\sin\theta on top — it is what writing v2=glv^{2} = gl at P gives, as though P were the top of the circle.

Q3JEE Main 2025One correct option

A bead of mass ' mm ' slides without friction on the wall of a vertical circular hoop of radius ' RR '. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ' RR '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' RR ', would be (spring constant is ' kk ', gg is accleration due to gravity)

  1. A2Rg+kR2m\sqrt{2Rg+\frac{{\mathrm{kR}}^{2}}{ m}}
  2. B3Rg+kR2m\sqrt{3\mathrm{Rg}+\frac{{\mathrm{kR}}^{2}}{ m}}
  3. C2Rg+4kR2m\sqrt{2\mathrm{Rg}+\frac{4{\mathrm{kR}}^{2}}{ m}}
  4. D2gR+kR2m2\sqrt{\mathrm{gR}+\frac{{\mathrm{kR}}^{2}}{ m}}
Show answer and solution

Answer: Option B

At the top the spring spans the whole diameter, 2R2R, so it is stretched by RR and stores 12kR2\tfrac{1}{2}kR^{2}. It is back to its natural length when the bead is a distance RR from the bottom point. That chord of length RR makes an equilateral triangle with the centre, so it subtends 60∘60^{\circ} there, and the bead is R(1−cos⁡60∘)=R2R(1 - \cos 60^{\circ}) = \dfrac{R}{2} above the bottom. So the bead has fallen from 2R2R to R2\dfrac{R}{2}, a drop of 3R2\dfrac{3R}{2}, and the spring has given up all its energy: 12mv2=12kR2+mg×3R2\tfrac{1}{2}mv^{2} = \tfrac{1}{2}kR^{2} + mg \times \dfrac{3R}{2}, so v=3gR+kR2mv = \sqrt{3gR + \dfrac{kR^{2}}{m}}. The hoop pushes along a radius, at right angles to the bead's motion, so it does no work. Option A takes the drop as RR, reading the spring's length as a height.

Practice questions, easy to hard

Three questions from the conservation of mechanical energy practice ladder: one easy, one medium, one hard.

Q4Numerical answer

A spring's potential energy 12kx2\tfrac{1}{2}kx^{2} belongs in the same total as mghmgh and 12mv2\tfrac{1}{2}mv^{2}. On a smooth level floor the height does not change, so a block running into a light spring simply trades its kinetic energy for spring energy, and the spring is squashed furthest at the instant the block stops.

A 2 kg2\ \mathrm{kg} block slides at 3 m/s3\ \mathrm{m/s} on a smooth floor into a light spring with k=200 N/mk = 200\ \mathrm{N/m}. What is the greatest compression, in m\mathrm{m}?

Show answer and solution

Answer: 0.3 m

At greatest compression the block is momentarily at rest, so all of its 12×2×32=9 J\tfrac{1}{2} \times 2 \times 3^{2} = 9\ \mathrm{J} is in the spring: 12×200×x2=9\tfrac{1}{2} \times 200 \times x^{2} = 9, so x2=0.09x^{2} = 0.09 and x=0.3 mx = 0.3\ \mathrm{m}. In general x=vm/kx = v\sqrt{m/k}. Setting kxkx equal to mvmv or to 12mv2\tfrac{1}{2}mv^{2} sets a force against a momentum or an energy — the balance has to be energy against energy, and the spring's energy carries its own half.

Q5One correct option

A loop-the-loop joins this chapter to the vertical circle. Energy gives the speed at the top of the loop; the circular-motion equation at the top, N+mg=mvtop2RN + mg = \dfrac{mv_{\mathrm{top}}^{2}}{R}, then gives the push of the track.

A small block is released from rest on a smooth track at a height 3R3R above the bottom of a vertical loop of radius RR. What force does the track exert on the block at the top of the loop?

  1. AZero
  2. Bmgmg
  3. C2mg2mg
  4. D3mg3mg
Show answer and solution

Answer: Option B

The top of the loop is 2R2R above the bottom, so the block has fallen 3R−2R=R3R - 2R = R by the time it gets there: vtop2=2gRv_{\mathrm{top}}^{2} = 2gR. Then N+mg=m×2gRR=2mgN + mg = \dfrac{m \times 2gR}{R} = 2mg, so N=mgN = mg, pressing down towards the centre. 2mg2mg forgets that gravity does part of the turning. Zero would need vtop2=gRv_{\mathrm{top}}^{2} = gR — the case of only just getting round, from a start at 2.5R2.5R.