Law of Conservation of Mechanical Energy — NEET Physics MCQs with Solutions
Free NEET Physics Law of Conservation of Mechanical Energy MCQs with step-by-step solutions (8 questions). Part of Work, Power and Energy. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Law of Conservation of Mechanical Energy · easy · theory
The total mechanical energy (KE + PE) of a system is conserved when:
A. Friction acts on it
B. Any force acts on it
C. Only conservative forces do work on it ✓ Correct
D. The system is at rest
Solution: If only conservative forces (gravity, spring…) do work, KE + PE = constant. Friction/air resistance drain mechanical energy into heat.
Q2 — Law of Conservation of Mechanical Energy · medium · theory
A ball falls freely from rest (no air resistance). During the fall:
A. Its PE stays constant
B. Both KE and PE increase
C. Its KE increases, PE decreases, and their sum stays constant ✓ Correct
D. Its total mechanical energy increases
Solution: Free fall converts PE to KE continuously; the sum mgh + ½mv² stays fixed at every instant.
Q3 — Law of Conservation of Mechanical Energy · easy · numerical
A ball is dropped from a height of 45 m. Its speed on reaching the ground is:
A. 45 m/s
B. 90 m/s
C. 21 m/s
D. 30 m/s ✓ Correct
Solution: mgh = ½mv² ⇒ v = √(2gh) = √(2×10×45) = √900 = 30 m/s.
Q4 — Law of Conservation of Mechanical Energy · medium · numerical
A stone is dropped from a height of 40 m. At what height above the ground are its kinetic and potential energies equal?
A. 25 m
B. 20 m ✓ Correct
C. 10 m
D. 30 m
Solution: KE = PE means each is half the initial PE ⇒ mgh = ½mg(40) ⇒ h = 20 m.
Q5 — Law of Conservation of Mechanical Energy · hard · numerical
A body is dropped from a height H = 20 m. Its speed when it has fallen through H/2 is:
A. 10√2 ≈ 14.1 m/s ✓ Correct
B. 14.1√2 m/s
C. 10 m/s
D. 20 m/s
Solution: v = √(2g·H/2) = √(gH) = √200 = 10√2 ≈ 14.1 m/s. Trap: half the height does NOT give half the final speed.
Q6 — Law of Conservation of Mechanical Energy · medium · numerical
A pendulum bob is released from a point 0.2 m above its lowest position. Its speed at the lowest point is:
A. 2 m/s ✓ Correct
B. 4 m/s
C. 1 m/s
D. 0.4 m/s
Solution: v = √(2gh) = √(2×10×0.2) = √4 = 2 m/s — the string tension does no work.
Q7 — Law of Conservation of Mechanical Energy · easy · numerical
A ball is thrown vertically downward at 10 m/s from a height of 15 m. Its speed on hitting the ground is:
A. 15 m/s
B. 25 m/s
C. 20 m/s ✓ Correct
D. 17.3 m/s
Solution: v² = u² + 2gh = 100 + 300 = 400 ⇒ v = 20 m/s (energy method: ½mv² = ½mu² + mgh).
Q8 — Law of Conservation of Mechanical Energy · medium · numerical
A ball thrown vertically up at 20 m/s has a speed of 10 m/s at height h. Then h equals:
A. 10 m
B. 15 m ✓ Correct
C. 5 m
D. 20 m
Solution: mgh = ½m(u² − v²) ⇒ h = (400 − 100)/(2×10) = 15 m.