The distinction between conservative and non-conservative forces decides whether you can use mechanical energy conservation — or whether you must account for energy lost to heat.
Conservative force (NCERT Class 11 Physics Chapter 5, page 8): a force where the work done on an object between two points is independent of the path taken. Equivalently, the work done around any closed path is zero. Gravity and the spring force are the standard examples. Because their work is path-independent, you can define a potential energy function for them.
Non-conservative force (NCERT Class 11 Physics Chapter 5, page 10): a force where the work done depends on the path. Friction is the canonical example — drag a block from A to B along a longer path and friction does more negative work. No potential energy function can be defined for friction.
Conservation of mechanical energy (NCERT Class 11 Physics Chapter 5, page 11): when only conservative forces act, K_i + U_i = K_f + U_f. Total mechanical energy is constant.
The high-frequency trap: aspirants apply K_i + U_i = K_f + U_f even when friction is explicitly present. The moment any non-conservative force does work, mechanical energy is NOT conserved. You must use the work-energy theorem instead: K_f − K_i = W_conservative + W_non-conservative, where the friction term is typically negative.
Watch-out: a problem that says "smooth surface" means no friction — conservation applies. A problem that says "rough surface" or gives a friction coefficient means non-conservative work is present — conservation does NOT apply in its simple form.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is a defining property of a conservative force?
Which of the following forces is non-conservative?
For a conservative force, which quantity can always be defined?
A 2 kg block slides down a smooth (frictionless) incline from a height of 5 m starting from rest. What is the speed of the block at the bottom? (Take g = 10 m/s²)
A 4 kg block slides down a rough incline through a vertical height of 3 m. The block's speed at the bottom is 6 m/s (starting from rest). How much energy was lost to friction? (Take g = 10 m/s²)
A block is pushed along a rough horizontal surface from point A to point B along a straight line (distance 3 m), then returned from B to A along the same straight line. The kinetic friction force on the block has magnitude 10 N throughout. What is the total work done by friction for the round trip?
A ball is thrown upward and returns to the thrower's hand. Air resistance is NOT negligible. Which statement about the ball's mechanical energy is correct?
A 1 kg block is released from rest at the top of a rough incline of height 10 m. Friction does −30 J of work on the block as it slides to the bottom. What is the speed of the block at the bottom? (Take g = 10 m/s²)
Given
A bob of mass m hangs on a string of length L. It is given a horizontal velocity v₀ at the lowest point. Take g as exact.
Required
Find the minimum v₀ for the bob to complete a full vertical circle without the string going slack.
Concept
Two conditions must be satisfied simultaneously: 1. Energy conservation (only gravity does work — string tension is always perpendicular to velocity, so it does zero work — conservative-force-only situation). 2. String tension T ≥ 0 at the topmost point (the string cannot push, only pull).
Formula
Energy conservation from bottom to top: ½mv₀² = ½mv_top² + mg(2L) Tension condition at top: mv_top²/L = mg + T, with T ≥ 0 → v_top² ≥ gL
Substitution
From energy: v_top² = v₀² − 4gL Applying the tension constraint: v₀² − 4gL ≥ gL
Calculation
v₀² ≥ 5gL Therefore v₀(min) = √(5gL) Note: g and the factor 5 are exact quantities (gravitational acceleration treated as exact in this problem; 5 is a counting integer). They do not limit significant figures.
Final answer
The minimum speed at the lowest point is v₀ = √(5gL).
Common trap
Using energy conservation alone gives v₀² ≥ 4gL (just enough KE to reach the top with zero speed). But zero speed at the top means zero centripetal acceleration and the string goes slack before reaching the top. The tension constraint adds the extra gL, giving 5gL. Energy alone is insufficient — this is a high-frequency error on this pattern.
Similar NEET-style question
A ball of mass 0.5 kg is whirled in a vertical circle on a string of length 2 m. Find the minimum speed at the lowest point for the ball to complete the circle. (Take g = 10 m/s².) Answer: v₀ = √(5 × 10 × 2) = √100 = 10 m/s. ---
Potential energy U(r) is energy possessed by a body by virtue of its position or configuration in a field of conservative force. The defining relation is F(r) = -dU(r)/dr (in 1D); U is defined up to an additive constant fixed by a chosen reference level.
-- NCERT Class 11 Physics, Ch. 5, p. 8A force is conservative if the work done by it on a particle moving between two points is independent of the path taken (equivalently, the work done on any closed loop is zero). Examples: gravity, spring force, electrostatic force. Non-conservative forces (friction, air drag) do path-dependent work; mechanical energy is not conserved when they act.
-- NCERT Class 11 Physics, Ch. 5, p. 10If 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. 11Final velocities of two bodies after an elastic head-on (1D) collision — momentum AND kinetic energy are both conserved. Special cases: equal masses exchange velocities; very heavy m2 at rest reflects m1 with reversed velocity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m1, m2 | Masses of the two bodies | kg |
| u1, u2 | Initial velocities (signed, before collision) | m/s |
| v1, v2 | Final velocities (signed, after collision) | m/s |
Potential energy of a body of mass m at height h above the chosen reference level, in a region where g is approximately constant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| U | Gravitational PE | J |
| m | Mass | kg |
| g | Gravitational acceleration | m/s^2 |
| h | Height above reference level | m |
Energy a body possesses by virtue of its motion. Always non-negative. Frame-dependent through v.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K | Kinetic energy | J |
| m | Mass | kg |
| v | Speed | m/s |
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 |
When two bodies stick together after a 1D collision, the common velocity is given by momentum conservation. The KE lost is converted to internal energy (heat, deformation).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Common final velocity | m/s |
| m1, m2 | Masses | kg |
| u1, u2 | Initial velocities | m/s |
| Delta_K | Change in kinetic energy | J |
Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Power (instantaneous) | W |
| F | Force | N |
| v | Velocity | m/s |
Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).
| Symbol | Quantity | SI Unit |
|---|---|---|
| U | Elastic PE | J |
| k | Spring constant (force per unit displacement) | N/m |
| x | Displacement from natural length | m |
The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W | Work (scalar, signed) | J |
| F | Constant force (vector) | N |
| s | Displacement (vector) | m |
| theta | Angle between F and s | rad/deg |
The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W_net | Net work done by all forces | J |
| Delta_K | Change in kinetic energy | J |
| m | Mass of particle | kg |
| v, v0 | Final, initial speed | m/s |
When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W | Work done | J |
| F(x) | Force as a function of position | N |
| dx | Infinitesimal displacement | m |
| x_i, x_f | Initial and final positions | m |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.
Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.
Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.
Category: Sign Convention
Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.
Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.
On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.
Category: Similar Terms
Student uses the elastic-collision velocity formulas (m1-m2)/(m1+m2) when the question explicitly says 'completely inelastic' (bodies stick).
Question says 'inelastic', 'stick together', 'after collision moves with common velocity'.
Perfectly inelastic 1D: v_common = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). KE is NOT conserved; loss = ½ (m₁ m₂)/(m₁+m₂) × (u₁ - u₂)². Don't use elastic formulas.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
Category: Overthinking
Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.
Question gives turbine, motor, or transformer with stated efficiency or loss percentage.
Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.
Category: Similar Terms
Student plugs displacement x into P = F·v formula instead of velocity v.
Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.
P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.
Category: Overthinking
Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.
Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).
U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².
Category: Overthinking
Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.
Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.
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: concept gap
Elastic: BOTH momentum and kinetic energy conserved -> use the (m1-m2)/(m1+m2) form. Inelastic: ONLY momentum conserved; KE generally lost to heat. Perfectly inelastic: bodies stick together -> common velocity = (m1*u1 + m2*u2)/(m1+m2). Identify which type from the problem before choosing a formula.
Distractor uses elastic-collision formulas for two bodies that the problem says stick together after impact.
Root cause: formula misuse
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).
Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.
Root cause: concept gap
PE is defined up to an additive constant. Only DIFFERENCES in PE have physical meaning (they equal the negative work done by the conservative force between two points). Choosing the ground as zero is a convention, not a derivation.
Distractor offers a numerical PE value where two different reference levels would give different correct numbers.
Root cause: direction ignored
Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.
Distractor reports W = N * s for a block sliding on a level surface.
ignores mu cos theta term
Forgetting the friction-along-incline component
forgets loss factor
Default to ideal-case formula
uses x instead of v
Default plugging x into the wrong formula
forgets friction term
Ignoring opposing force
no KE loss stated
Treats collision as elastic by default
linear scaling
Thinking PE scales as x not x^2
uses only energy not tension
Energy gives speed but tension constraint sets the floor
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