If only conservative forces do work on a system, the total mechanical energy E = K + U remains constant: K_i + U_i = K_f + U_f. When non-conservative forces do work, the change in mechanical energy equals the work done by those non-conservative forces (typically negative for friction/drag).
-- NCERT Class 11 Physics, Ch. 5, p. 78Conservation Mechanical Energy
8 MCQs3 revision cards8-step worked example
Source: NCERT Work, Energy and PowerOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Conservation Mechanical Energy, explained for NEET
The costly habit on this topic is reaching for K_i + U_i = K_f + U_f the moment a height appears in the problem. The equation has a precondition, and the precondition is not "gravity is involved" — it is that only conservative forces do work. A block sliding down a rough incline has a height drop and a speed at the bottom, and mgh = ½mv² is wrong there, because friction removed energy from the mechanical account. Write the work–energy theorem instead: K_f − K_i = W_conservative + W_non-conservative, with the friction term negative.
NCERT Class 11 Physics Chapter 5 states the law on page 81: when the forces doing work are conservative, the total mechanical energy E = K + U of the system stays constant in time. Page 82 gives the working form, K_i + U_i = K_f + U_f. Two conditions travel with it — conservative forces only, and a closed system with no external energy input.
Notice what the law does not promise. It says nothing about the path taken, nothing about how long the journey lasts, and nothing about the forces present that do no work. A normal reaction perpendicular to the motion, or a string tension on a bob swinging in a circle, does zero work — both are allowed in a conservation problem even though neither is a conservative force. The test is always "does this force do work?", never "is this force conservative?".
For NEET this appears in two guises: (i) a direct energy balance on a smooth track or a free fall, where the answer follows in one line; (ii) a vertical-circle problem, where energy conservation gives you the speed at the top but is insufficient on its own — the string must also stay taut, which requires v_top² ≥ gL. Energy alone yields v₀² ≥ 4gL; the correct minimum is v₀² ≥ 5gL. That gap between 4 and 5 is the whole question.
Watch-out: before writing the conservation equation, list every force and ask which ones do work. If any non-conservative one survives that filter, the equation does not apply.
Can you answer these Conservation Mechanical Energy MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The conservation of mechanical energy, as stated in NCERT Class 11 Physics Chapter 5, holds for a system when:
Show answer and why every option is right or wrong
Answer: C. The law as stated on page 81 of NCERT Class 11 Physics Chapter 5 attaches exactly one condition — that the forces doing work are conservative. Then E = K + U is constant.
Why A is wrong: A is wrong because zero net force is the condition for momentum conservation, not mechanical-energy conservation. A block sliding at constant velocity on a rough floor has zero net force while friction steadily drains mechanical energy.
Why B is wrong: B is wrong because a closed path guarantees zero work only for a conservative force. Friction around a closed loop does non-zero negative work, so mechanical energy is not conserved on a round trip over a rough surface.
Why D is wrong: D is wrong because constant kinetic energy is neither required nor implied. In free fall K rises continuously while U falls by the same amount — that is precisely the law in action.
MCQ 2Concept TrapPractice
A block slides from rest down a rough incline and reaches the bottom after a vertical drop h. A student writes mgh = ½mv² and solves for v. The result is:
Show answer and why every option is right or wrong
Answer: D. Friction is non-conservative and does negative work, so the true kinetic energy at the bottom is less than mgh. Equating the two credits the block with energy that went to heat, giving an overestimate — the condition on page 81 of NCERT Class 11 Physics Chapter 5 is violated.
Why A is wrong: A is wrong because the statement about the normal reaction is true but irrelevant. Zero work by N does not license the equation while a second force, friction, is doing negative work.
Why B is wrong: B is wrong because it confuses a true statement about U with the applicability of the whole equation. The drop in U is indeed mgh, but that energy splits between kinetic energy and heat, so it does not all appear as ½mv².
Why C is wrong: C is wrong on the direction of the friction force. Friction on a block sliding down opposes the motion and so opposes the driving component, reducing the speed at the bottom, not increasing it.
MCQ 3Direct ApplicationPractice
A bob of mass m hangs on a light inextensible string of length L and is given a horizontal speed u at the lowest point. Assuming the string stays taut and air resistance is negligible, the speed of the bob when the string has swung to the horizontal position is:
Show answer and why every option is right or wrong
Answer: C. The bob rises by L reaching the horizontal, so ½mu² = ½mv² + mgL, giving v² = u² − 2gL. Tension does no work because it stays perpendicular to the velocity, so the conservation form on page 82 of NCERT Class 11 Physics Chapter 5 applies directly.
Why A is wrong: A is wrong because 4gL corresponds to a rise of 2L, which is the height of the topmost point of the circle, not the horizontal position. The horizontal position is a rise of L only.
Why B is wrong: B is wrong because it uses a rise of L/2. Measuring from the lowest point, the string's horizontal position is a full string-length above it.
Why D is wrong: D is wrong on the sign. The bob is rising, so potential energy increases and kinetic energy must fall; adding 2gL would mean the bob speeds up as it climbs.
MCQ 4Concept TrapPractice
For the swinging bob of the previous set-up, the string tension is not a conservative force, yet mechanical energy is still conserved. The reason is that:
Show answer and why every option is right or wrong
Answer: B. The conservation condition is about which forces do work, not about which forces are conservative. The velocity of the bob is tangential and the tension is radial, so W = Fs cos 90° = 0 and the tension never enters the energy account — consistent with the statement on page 81 of NCERT Class 11 Physics Chapter 5.
Why A is wrong: A is wrong as physics: the tension exceeds the radial weight component by mv²/L to supply the centripetal acceleration. Even were it true, equal magnitudes would not make the work zero.
Why C is wrong: C is wrong because being internal is not the criterion. Internal non-conservative forces — friction between two blocks in contact, for instance — destroy mechanical energy just as external ones do.
Why D is wrong: D is wrong because inextensibility does not make a force conservative. There is no potential-energy function associated with string tension; the reason it drops out is geometric, not a property of the string.
MCQ 5CalculationPractice
A bob of mass m on a light string of length L is given a horizontal speed v₀ at the lowest point. For the string to remain taut throughout a complete vertical circle, the minimum required value of v₀² is:
Show answer and why every option is right or wrong
Answer: B. Two conditions must both hold. Energy conservation between lowest and topmost points gives v_top² = v₀² − 4gL; the string stays taut at the top only if the weight can supply no more than the required centripetal force, i.e. v_top² ≥ gL. Combining, v₀² ≥ 5gL.
Why A is wrong: A is wrong because 2gL is the energy needed to rise by L — a quarter-circle to the horizontal. The bob would not even reach the top.
Why C is wrong: C is wrong in exactly the way this pattern's distractor is designed to catch: it uses the energy condition v₀² ≥ 4gL alone, which only makes v_top reach zero. A bob arriving at the top with zero speed needs zero centripetal force, but gravity still supplies mg — the string has already gone slack.
Why D is wrong: D is wrong because 6gL is neither condition. It arises from adding gL to 5gL a second time, or from imposing v_top² ≥ 2gL with no justification; the tension floor is v_top² = gL, from mg = mv_top²/L.
MCQ 6Direct ApplicationPractice
A body is projected vertically upward from the ground with speed u. Neglecting air resistance, the height at which its kinetic energy equals its gravitational potential energy measured from the ground is:
Show answer and why every option is right or wrong
Answer: A. Total mechanical energy is ½mu² throughout. Setting K = U makes each equal to half the total, so mgh = ¼mu² and h = u²/(4g) — a one-line application of the form on page 82 of NCERT Class 11 Physics Chapter 5.
Why B is wrong: B is wrong because u²/(2g) is the maximum height, where K = 0 and U carries the entire energy. The equal-split point sits halfway up.
Why C is wrong: C is wrong because it sets mgh = ⅛mu², halving the energy share one time too many. Each of K and U takes half of ½mu², which is ¼mu², not ⅛mu².
Why D is wrong: D is wrong because it is twice the maximum height and so unreachable. Any answer for this body exceeding u²/(2g) can be rejected without calculation.
MCQ 7Easy RecallPractice
Which of these situations permits the direct use of K_i + U_i = K_f + U_f?
Show answer and why every option is right or wrong
Answer: D. On a frictionless wire the only forces are gravity (conservative) and the normal reaction from the wire, which is perpendicular to the motion and does no work. The shape of the wire is irrelevant — the conditions listed with the law on page 81 of NCERT Class 11 Physics Chapter 5 are met.
Why A is wrong: A is wrong because air drag is non-conservative and is doing large negative work; that is precisely why the speed is constant while the parachutist keeps descending. Mechanical energy falls steadily.
Why B is wrong: B is wrong because braking converts kinetic energy into heat at the brakes and tyres. Non-conservative work is the entire content of the process.
Why C is wrong: C is wrong because a bullet embedding in a block is a perfectly inelastic event: kinetic energy is lost to deformation and heat, so mechanical energy is not conserved even though momentum is.
MCQ 8CalculationPractice
A small block is released from rest at the top of a smooth hemispherical dome of radius R and slides down its outer surface. Using energy conservation together with the requirement that the normal reaction cannot be negative, the height above the centre of the dome at which the block leaves the surface is:
Show answer and why every option is right or wrong
Answer: A. With the block at angular position θ from the top, energy conservation gives v² = 2gR(1 − cos θ). The block leaves when N = 0, i.e. mg cos θ = mv²/R, giving cos θ = 2/3 and a height R cos θ = 2R/3.
Why B is wrong: B is wrong because it comes from using v² = gR(1 − cos θ), dropping the factor of 2 in the energy equation; N = 0 then gives cos θ = 1 − cos θ, so cos θ = 1/2. ½mv² = mgR(1 − cos θ) carries that 2 through and fixes cos θ = 2/3.
Why C is wrong: C is wrong because a height R above the centre is the very top of the dome. The block starts there from rest, where N = mg > 0, so it stays on the surface and gains speed before it leaves.
Why D is wrong: D is wrong because R/3 is the drop below the top, not the height above the centre. The two are complements: R − 2R/3 = R/3, so this option reports the correct physics against the wrong reference point.
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Conservation Mechanical Energy: quick recall before you leave
How do you solve a Conservation Mechanical Energy question? A worked example
- 1
Given.
• Mass
m = 2.0 × 10⁻¹ kg• String lengthL = 4.0 × 10⁻¹ m•g = 10 m/s²(exact, problem-defined)• The lowest-point speedv₀is the minimum for the string to stay taut over the full circle. - 2
Required.
v₀(minimum speed at lowest point) andv_top(speed at the highest point). - 3
Concept.
Two independent conditions govern a vertical circle on a string. The string can pull but cannot push, so the tension at the top must be at least zero — a dynamics condition. Separately, the bob must have enough energy to reach the top — an energy condition, supplied by mechanical-energy conservation, valid here because tension does no work (it is perpendicular to the velocity at every instant) and gravity is conservative.
4. Formulas.• Conservation of mechanical energy:K_i + U_i = K_f + U_f• Kinetic energy:K = ½mv²• Gravitational PE:U = mgh• Centripetal condition at the top withT = 0:mg = mv_top²/L - 5
Substitution.
Take the lowest point as the reference level, so
U_i = 0. The highest point is a height2Labove it.
Tension condition at the top (limiting caseT = 0):mg = m v_top² / L⟹v_top² = gL
Energy condition, lowest to highest:½ m v₀² + 0 = ½ m v_top² + m g (2L) - 6
Calculation.
The mass
mcancels throughout — worth noting, since the answer is independent of the bob's mass.v_top² = gL = 10 × 4.0 × 10⁻¹ = 4.0 m²/s²v_top = 2.0 m/sv₀² = v_top² + 4gL = gL + 4gL = 5gL = 5 × 10 × 4.0 × 10⁻¹ = 2.0 × 10¹ m²/s²v₀ = √20 = 4.472... m/s
Theg = 10 m/s²is problem-defined as exact and the5in5gLis a counting factor from the algebra, so neither limits precision. The significant-figure count is set by the two measured quantities,L = 4.0 × 10⁻¹ m(2 s.f.) andm = 2.0 × 10⁻¹ kg(2 s.f., and it cancels anyway). Two significant figures are carried to the answer. - 7
Final answer.
• Minimum speed at the lowest point:
v₀ = 4.5 m/s(2 s.f.)• Speed at the highest point:v_top = 2.0 m/s(2 s.f.) - 8
Common trap.
The energy equation alone gives
v₀² ≥ 4gL, i.e.v₀ ≥ 4.0 m/s— the condition for the bob merely to arrive at the top. That answer is wrong, and it is wrong in a way that looks complete: the algebra is clean and the number is plausible. A bob arriving at the top with vanishing speed needs vanishing centripetal force, but gravity is still pulling down withmg, so the string went slack before it got there. The tension floorv_top² ≥ gLmust be imposed as well, lifting the requirement from4gLto5gL. When a problem involves a string or an inner track, energy conservation is necessary but never sufficient. - 9
Similar NEET-style question.
A particle of mass
mis attached to a light rod of lengthL— a rigid rod, not a string — and rotated in a vertical circle. What is the minimum speed at the lowest point for the particle to complete the circle? *(A rod can push as well as pull, so the tension floor disappears and only the energy condition survives:v₀² ≥ 4gL. The string-versus-rod swap is the standard twist on this pattern, and the5gLanswer is the trap in that version.)*
What to remember before solving Conservation Mechanical Energy questions
For a body of mass m tied to a string of length r and rotating in a vertical circle, the minimum speed at the topmost point (for the string to remain taut) is v_top = √(g r). From energy conservation, the corresponding speed at the bottom is v_bottom = √(5 g r).
-- NCERT Class 11 Physics, Ch. 5, p. 79Which Conservation Mechanical Energy formulas do you need for NEET?
1 formula — click to collapse
Conservation of mechanical energy
If only conservative forces do work on a system, the total mechanical energy E = K + U is constant in time.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_i, K_f | Initial, final kinetic energy | J |
| U_i, U_f | Initial, final potential energy | J |
Valid when
- Only CONSERVATIVE forces do work (gravity, spring, electrostatic — not friction or drag)
- Closed system; no energy exchange with surroundings
Do NOT use when
- Friction, drag, or other non-conservative forces do work
- External forces add energy to the system
Where do students lose marks on Conservation Mechanical Energy?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Overthinking
Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.
When it triggers
Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.
How to avoid
TWO constraints: (1) energy: v_top² = v₀² - 4gL; (2) tension at top ≥ 0: v_top² ≥ gL. Combined: v₀² ≥ 5gL. Energy alone gives only v₀² ≥ 4gL which is insufficient.
Root cause: formula misuse
Correction
Mechanical-energy conservation requires that ONLY conservative forces do work. When friction or drag is present, use the work-energy theorem directly: K_f - K_i = W_conservative + W_non-conservative, where W_non-conservative is typically negative (energy goes to heat).
Wrong option pattern
Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.
More in Work, Energy and Power: 10 exam traps and mistakes · 9 formulas · 6 question patterns from its other lessons.
Conservation Mechanical Energy questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Conservation Mechanical Energy?
1 recurring pattern from past papers — click to collapse
Bob of mass m on string of length l given horizontal velocity v₀ at lowest point; find condition for string to remain taut throughout the vertical circle. Minimum: v₀² ≥ 5gl. At top: v_top² = v₀² - 4gl ≥ gl. Energy conservation + tension at any angle.
Multi StepHard
Common distractors
uses only energy not tension
Energy gives speed but tension constraint sets the floor
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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