Q.A body of mass 1 kg is thrown upwards with a velocity 20 ms−1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (Take g=10 ms−2) (AIPMT 2009)
Concept understanding — Conservation Of Mechanical Energy
Conservation of Mechanical Energy
The Intuition First
Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.
Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.
This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.
The Precise Statement
Conservation of Mechanical Energy: In an isolated system where only conservative forces (like gravity or an ideal spring) do work, the total mechanical energy of the system remains constant.
Total mechanical energy is the sum of kinetic energy (K) and potential energy (U):
Emech=K+U
The law says:
Kinitial+Uinitial=Kfinal+Ufinal
Or, in symbols:
Emech, initial=Emech, final
What This Means in Practice
Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take g=10 m/s2.
-
At the top (initial):
Ki=0 (not moving)
Ui=mgh=2×10×5=100 J
Emech=0+100=100 J
-
Just before hitting ground (final):
Uf=0 (height = 0)
Kf=21mv2
Conservation says Kf=100 J, so 21×2×v2=100, giving v=10 m/s.
You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.
The Two Critical Conditions
Mechanical energy is not always conserved. It is conserved only when:
- No non-conservative forces (like friction, air resistance, or applied pushes/pulls) do work.
- The system is isolated — no external forces transfer energy in or out.
If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.
A Simple Example to Cement It
A pendulum swings. At the highest point on either side, it stops momentarily — all energy is potential. At the lowest point, it moves fastest — all energy is kinetic. In between, it's a mix. If there were no air resistance or friction at the pivot, the pendulum would swing forever, with the same maximum height each time. That's mechanical energy being perfectly conserved.
The Big Picture
Conservation of mechanical energy is a shortcut. Instead of analysing forces and accelerations (which can be messy), you just track two numbers — kinetic and potential — and set their sum equal at two moments. It works because energy is a scalar (no direction), so you don't need vectors or free-body diagrams.
K1+U1=K2+U2
where K=21mv2 and U depends on the force (e.g., U=mgh for gravity, U=21kx2 for a spring).
Final takeaway: Mechanical energy is never created or destroyed — only converted between kinetic and potential forms, as long as no friction or other non-conservative forces interfere.
Searches for "Conservation Of Mechanical Energy notes class 11" and "Conservation Of Mechanical Energy important questions" both point back to this same core idea, since Conservation Of Mechanical Energy is a syllabus-aligned topic under Work, Energy and Power in NCERT Class 11 Physics, making it a natural fit for both board exams and JEE/NEET practice sets. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.
The work-energy theorem gives the energy lost to friction as the initial kinetic energy minus the gravitational potential energy actually gained.
(a) 20 J
Step 1. Initial kinetic energy, KEi=21mu2=21(1)(20)2=200 J.
Step 2. The body rises to height h=18 m before momentarily stopping, so its final kinetic energy is zero and the potential energy gained is U=mgh=(1)(10)(18)=180 J.
Step 3. By the work-energy theorem, KEi+Wgravity+Wfriction=KEf: 200+(−180)+Wfriction=0, so Wfriction=−20 J.
Step 4. The magnitude of energy lost to air friction is therefore 20 J -- exactly the shortfall between the height the body would have reached without friction (u2/2g=400/20=20 m) and the height it actually reached (18 m), scaled by mg: (20−18)×1×10=20 J, confirming the answer.
(a) 20 J.
Apply the work-energy theorem: initial KE = final PE gained + energy lost to friction (since final KE = 0).
- Computing the ideal frictionless height (20 m) and stopping there instead of comparing it against the actual height reached (18 m).
- Forgetting that the final kinetic energy is zero at the highest point, not the initial kinetic energy.
- CBSE 2022Set TERM11 markMCQQ.A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is(1) 316 J(2) 216 J(3) 116 J(4) 392 J
›Reveal solutionSolution
Total mechanical energy = kinetic energy + potential energy. Computing each with m=10 kg, v=2 m/s, h=2 m, g=9.8 m/s^2 and adding gives 216 J.
Given: mass m = 10 kg, height h = 2 m, speed v = 2 m/s, g = 9.8 m/s^2.
Kinetic energy:
KE = (1/2) m v^2 = (1/2)(10)(2^2) = (1/2)(10)(4) = 20 J
Potential energy (relative to the ground):
PE = m g h = (10)(9.8)(2) = 196 J
Total mechanical energy:
E = KE + PE = 20 + 196 = 216 J
✓Final answer(2) 216 J.
- CBSE 2022Set ANNUAL1 markMCQQ.The difference of tension between lowest point & highest point is —(a) 6 mg(b) 3 mg(c) mg(d) None of these
›Reveal solutionSolution
The difference in tension between the lowest and highest points of a vertical circle is 6mg.
At the lowest point: TL−mg=rmvL2, so TL=mg+rmvL2.
At the highest point: TH+mg=rmvH2, so TH=rmvH2−mg.
Subtracting: TL−TH=2mg+rm(vL2−vH2).
By energy conservation between top and bottom (height difference 2r): vL2−vH2=2g(2r)=4gr.
So TL−TH=2mg+rm(4gr)=2mg+4mg=6mg.
✓Final answer(a) 6 mg.
- CBSE 2019Set ANNUAL1 markMCQQ.A body of mass 5 kg is thrown up vertically with a kinetic energy of 1000 J. If acceleration due to gravity is 10 ms^-2, find the height at which the kinetic energy becomes half of the original value.(a) 10 m(b) 20 m(c) 50 m(d) 100 m
›Reveal solutionSolution
The kinetic energy converted to potential energy as the body rises is 500 J (half of the initial 1000 J), giving a height of 10 m.
As the body of mass m = 5 kg moves upward, mechanical energy is conserved (ignoring air resistance): the kinetic energy it loses is converted into gravitational potential energy gained.
Initial KE = 1000 J. The KE becomes half its original value, i.e. KE_final = 500 J.
KE lost = 1000 - 500 = 500 J
This lost KE equals the potential energy gained:
mgh = 500
5 x 10 x h = 500
50h = 500
h = 10 m
✓Final answerThe correct option is (a) 10 m.
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