Conservation of mechanical energy

What is the conservation of mechanical energy?

In plain words: the total mechanical energy of a body is its kinetic energy plus its potential energy, E = K + V. As the body moves, kinetic energy can turn into potential energy and back again, but if only conservative forces such as gravity or a spring force do work, the total stays the same.

This equation applies when the potential energy has no explicit time dependence and the net work of forces not included in that potential is zero. More generally, the change in mechanical energy equals that additional work: Δ(K + V) = W_nc. Such work can add energy, as for a motor, or remove it, as for sliding friction on a fixed surface.

Why mechanical energy is conserved

Take a constant-mass particle in an inertial frame moving through an infinitesimal displacement dx under a time-independent conservative force F(x).

By the work-energy theorem, the change in kinetic energy is the work done:

Because the force is conservative, it has a potential energy V(x) whose change is the negative of the work done:

Adding the two equations:

So K + V does not change from point to point: its value at the start equals its value at the end. The kinetic and potential energies may each vary, but their sum is constant.

Three ways to recognise a conservative force

NCERT gives three equivalent definitions. A force F(x) is conservative if:

1. It comes from a potential energy, through F = −dV/dx (in three dimensions this needs a vector derivative, beyond Class 11).
2. Its work depends only on the end points, not on the path: W = V(x_i) − V(x_f).
3. Its work around any closed path is zero, since a closed path starts and ends at the same point.

Gravity and the ideal spring force pass all three tests. Friction and air resistance fail them: their work depends on the path, and they turn mechanical energy into heat. When such forces do work, mechanical energy is not conserved; the loss equals the work done against them.

A falling ball

Drop a ball of mass m from rest from a cliff of height H (taking the ground as the reference level for potential energy). Ignoring air resistance, its total mechanical energy is the same at every height:

Equating the totals gives the speed at the ground and at height h:

the same results as from the equations of motion for free fall. At the top the energy is all potential; on the way down it is partly kinetic; at the ground it is all kinetic.

Motion in a vertical circle

A bob of mass m on a light inextensible string of length L, attached to a fixed pivot and moving without air resistance, is given a horizontal speed v₀ at the lowest point A, so that it swings round a vertical circle. Only gravity does work: the tension is always at right angles to the bob’s motion. So mechanical energy is conserved. NCERT Example 5.7 (pages 79–80) finds the limiting circular motion for which tension just vanishes at the top, C. The bob remains on the circle; zero tension at that single instant does not mean that it leaves the circle.

At the top C, the tension is zero, so gravity alone supplies the centripetal force:

C is 2L above A, so the energy at C is

At the bottom A, all of this energy is kinetic:

At B, level with the centre, a height L above A:

so the ratio of kinetic energies is

The tensions follow from the second law towards the centre: the net inward force is the centripetal force mv^²/L. At A, T_A − mg = mv₀²/L, so T_A = 6mg. At B, the tension alone points to the centre: T_B = mv_B²/L = 3mg. At C, T_C = 0. NCERT writes the equation at A but does not solve it; these values are a standard extension. In general, as long as the string stays taut, T_A − T_C = 6mg for any starting speed; T_A = 6mg only when T_C = 0.

After C the bob keeps moving round the circle. As NCERT notes, if the string were cut at C, the bob would move off horizontally to the left as a projectile.

Two more standard results go beyond NCERT’s example:

• If v₀ is between $\sqrt{2gL}$ and $\sqrt{5gL}$, the string goes slack somewhere above the level of O, at a height (v₀² + gL)/(3g) above A, and the bob leaves the circle as a projectile. If $v_0 \le \sqrt{2gL}$, the bob never rises above the level of O and the string never goes slack: it just swings to and fro.
• With a light rod instead of a string, the rod can push as well as pull, so the bob only needs to reach the top with (just over) zero speed: a speed just over $\sqrt{4gL}$ at the bottom is enough.

Energy conservation around us

• A swinging pendulum. Its energy changes from potential at the ends of the swing to kinetic at the bottom, and back. Air resistance slowly takes energy away, so the swing dies down.
• Roller coasters. The car is pulled to the top once; after that, its speed at any point follows from the height it has fallen, minus small losses to friction.
• Hydroelectric power. Water falling from a dam turns potential energy into kinetic energy, which the turbines then convert into electrical energy.
• A bouncing ball. Each bounce is lower because some energy is lost as heat and sound in the collision: mechanical energy is not conserved there.

Solved examples

Common mistakes to avoid

Quick revision summary

• Total mechanical energy is kinetic plus potential energy, E = K + V.
• If only conservative forces do work, it stays constant: the kinetic and potential energies trade off but their sum does not change.
• A force is conservative if it comes from a potential energy, if its work depends only on the end points, and if its work around a closed path is zero.
• For a body falling from rest through H, the speed is √(2gH); at height h it is √(2g(H − h)).
• In a vertical circle on a string, the bob just completes the circle with √(5gL) at the bottom, √(3gL) level with the centre and √(gL) at the top, where the tension is zero.

Questions students ask

What is the difference between conservation of energy and conservation of mechanical energy?

Conservation of energy says that the total energy of an isolated system, in all its forms, never changes. Conservation of mechanical energy is narrower: for a time-independent potential, kinetic plus potential energy stays constant when net non-conservative work is zero. Dissipative sliding friction on a fixed surface converts mechanical energy into internal energy. Other non-conservative forces, such as a motor’s drive, can increase mechanical energy while total energy remains conserved.

What is the minimum speed at the bottom needed to complete a vertical circle?

The minimum speed at the lowest point for a bob on a light string of length L to complete a vertical circle is √(5gL). At that limiting speed the tension just vanishes at the top, where the bob moves at √(gL) and its weight alone supplies the centripetal force, and energy conservation over the height 2L gives v₀² = gL + 4gL = 5gL. Between √(2gL) and √(5gL) the string goes slack above the level of the centre and the bob leaves the circle; below √(2gL) it only swings to and fro.

Why is the tension at the bottom of a vertical circle 6mg more than at the top?

The tension at the bottom of a vertical circle is 6mg more than at the top because the bob is faster at the bottom and gravity pulls against the tension there but with it at the top. At the bottom A, T_A − mg = mv₀²/L, and at the top C, T_C + mg = mv_C²/L, so T_A − T_C = 2mg + m(v₀² − v_C²)/L. Energy conservation over the height 2L gives v₀² − v_C² = 4gL, so T_A − T_C = 6mg at any speed, as long as the string stays taut.

Is motion in a vertical circle in the JEE Main and NEET syllabus?

Yes, motion in a vertical circle is in both syllabuses: the JEE Main 2027 syllabus and the NEET (UG) 2026 syllabus list it in the Work, Energy and Power unit, alongside conservative and non-conservative forces. In NCERT it is Example 5.7 of the Class 11 chapter Work, Energy and Power, solved with conservation of mechanical energy.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A roller coaster car starts from rest at a point 30 m high. Ignoring friction, find its speed at a point 10 m high. Take g = 9.8 m s⁻².

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   It has fallen 20 m:

   $$\begin{aligned} v &= \sqrt{2 \times 9.8 \times 20} \\ &\approx 19.8\ \text{m s}^{-1} \end{aligned}$$

2. Q2. A pendulum 2 m long is pulled aside to 30° from the vertical and released. Find its speed at the lowest point. Take g = 9.8 m s⁻².

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   It falls through h = 2(1 − cos 30°), about 0.268 m, so

   $$\begin{aligned} v &= \sqrt{2 \times 9.8 \times 0.268} \\ &\approx 2.29\ \text{m s}^{-1} \end{aligned}$$

3. Q3. What is the least speed a bob on a string 1.2 m long must have at the lowest point to complete a vertical circle? Take g = 9.8 m s⁻².

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   $$\begin{aligned} v_0 &= \sqrt{5gL} \\ &= \sqrt{5 \times 9.8 \times 1.2} \\ &= \sqrt{58.8} \approx 7.67\ \text{m s}^{-1} \end{aligned}$$