Energy Conservation in Free Fall and Falling Bodies
Physics · Work, Energy And Power · NEET
When a body falls freely, its gravitational potential energy (mgh) turns into kinetic energy (1/2 mv^2), but the total mechanical energy stays constant because gravity is a conservative force and air resistance is ignored. So at any point, PE + KE = mgH (the starting energy), and the speed after falling a height h is v = sqrt(2gh). Memory hook: "PE spends, KE earns, the total never burns."
As a body falls from height H, potential energy (red line) drops and kinetic energy (blue line) rises by the same amount, so their sum stays fixed at mgH (purple dashed line). KE is maximum at the ground where v = sqrt(2gH).
Your doubts, answered
Does a heavier body reach the ground with more speed in free fall?
No. In the energy equation mgh = 1/2 mv^2, the mass m cancels on both sides, giving v = sqrt(2gh). The final speed depends only on the height h and g, not on mass. A 1 kg stone and a 5 kg stone dropped from the same height hit the ground at the same speed (in vacuum). The heavier body has more energy, but also needs more energy to reach that same speed, so the two effects cancel.
Why does total mechanical energy stay constant during free fall?
Because the only force doing work is gravity, which is a conservative force. A conservative force gives back exactly the energy it takes: as the body falls, PE decreases and KE increases by the same amount, so their sum PE + KE never changes. This is only true when air resistance (a non-conservative force) is neglected.
Is energy still conserved when air resistance acts on the falling body?
Mechanical energy is NOT conserved. Air resistance is non-conservative and does negative work, so some mechanical energy is lost as heat and sound. Then use the work-energy theorem: W_gravity + W_air = change in KE. Total energy of the universe is still conserved, but PE + KE of the body alone decreases. The NEET rain-drop PYQ tests exactly this idea.
Where is kinetic energy maximum and potential energy minimum during a fall?
KE is maximum and PE is minimum at the lowest point (just before hitting the ground). At the start (top), the body is at rest, so KE = 0 and PE = mgH is maximum. As it falls, PE steadily converts to KE. At the ground (taking that as reference), PE = 0 and KE = mgH.
When should I use energy conservation instead of v = sqrt(2gh)?
Use energy conservation (mgH = mgh + 1/2 mv^2) whenever a question mixes potential and kinetic energy, asks for the fraction of energy at some height, or involves a curved or frictionless track where kinematics gets messy. v = sqrt(2gh) is just the special case for a straight vertical free fall from rest. Energy conservation is more general and faster for track and pendulum problems.
⚠️ The NEET trap ✗ Assuming the final speed depends on mass, or forgetting air resistance so you equate mgh directly to 1/2 mv^2 when the drop clearly slows down. ✓ For pure free fall in vacuum, v = sqrt(2gh) is mass-independent. But if air resistance is mentioned, mechanical energy is not conserved: use W_gravity + W_air = change in KE, where W_air is negative. 🧠 See 'air resistance' or 'resistive force'? Do NOT use energy conservation alone. Switch to the work-energy theorem.
Real NEET questions
NEET 2021
A particle is released from height S from the surface of the Earth. At a 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:
A · 3S/4, sqrt(3gS/2)
B · S/4, sqrt(3gS/2) ✓
C · S/4, 3gS/2
D · 3S/4, 3gS/2
Solution: Take the ground as PE = 0. Total energy at release = mgS (all potential, since released from rest). At the height h where KE = 3 PE: total energy = KE + PE = 3PE + PE = 4PE = 4mgh. By conservation, 4mgh = mgS, so h = S/4. Now KE = 3 PE = 3 mg(S/4) = 3mgS/4. Set equal to 1/2 mv^2: 1/2 v^2 = 3gS/4, so v^2 = 3gS/2, giving v = sqrt(3gS/2). Answer: h = S/4 and v = sqrt(3gS/2), option B.
NEET 2017
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g = 10 m/s^2. The work done by the (i) gravitational force and the (ii) resistive force of air is:
A · (i) 1.25 J (ii) -8.25 J
B · (i) 100 J (ii) 8.75 J
C · (i) -10 J (ii) -8.25 J
D · (i) 10 J (ii) -8.75 J ✓
Solution: Mass m = 1 g = 10^-3 kg, height h = 1 km = 10^3 m. Work by gravity W_grav = mgh = 10^-3 x 10 x 10^3 = 10 J. Change in kinetic energy = 1/2 m v^2 = 1/2 x 10^-3 x (50)^2 = 1/2 x 10^-3 x 2500 = 1.25 J. By the work-energy theorem, W_grav + W_air = change in KE, so W_air = 1.25 - 10 = -8.75 J. The negative sign shows air resistance removes mechanical energy. Answer: (i) 10 J, (ii) -8.75 J, option D.
NEET 2026
The sum of the kinetic energy and potential energy of a simple pendulum bob is 0.02 J. The speed of the bob at the equilibrium position is approximately (mass of the bob = 20 g):
A · 0.2 m/s
B · 1.41 m/s ✓
C · 14.1 m/s
D · 2.0 m/s
Solution: The total mechanical energy PE + KE = 0.02 J is constant (energy conservation, gravity does the work and tension does none). At the equilibrium (lowest) position, PE is taken as zero, so all energy is kinetic: 1/2 m v^2 = 0.02 J. Mass m = 20 g = 0.02 kg. Then v^2 = 2(0.02)/0.02 = 2, so v = sqrt(2) = 1.41 m/s. Answer: option B.
Solved Work, Energy And Power NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for speed of a freely falling body using energy conservation?
Starting from rest and falling a height h, mgh = 1/2 mv^2, so v = sqrt(2gh). The mass cancels, so speed depends only on height and g.
Is mechanical energy conserved in free fall?
Yes, if air resistance is neglected. Gravity is a conservative force, so PE + KE stays constant. If air resistance acts, mechanical energy is not conserved and you must use the work-energy theorem.
At the top of a free fall, what is the kinetic energy?
If the body is released from rest, its kinetic energy at the top is zero and its potential energy mgH is maximum. As it falls, this PE converts fully into KE.
Why does the mass not affect the final speed in free fall?
In mgh = 1/2 mv^2 the mass m appears on both sides and cancels, leaving v = sqrt(2gh). So all bodies fall at the same rate in vacuum, regardless of mass.
How is energy conservation used in a vertical circle or frictionless track problem?
Write total energy = KE + PE at two points and set them equal (mgh1 + 1/2 mv1^2 = mgh2 + 1/2 mv2^2). This avoids kinematics along curved paths and directly links height changes to speed changes.
