Work Energy Theorem — NEET Physics MCQs with Solutions
Free NEET Physics Work Energy Theorem 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 — Work Energy Theorem · easy · theory
The work–energy theorem states that the NET work done on a particle equals:
A. The change in its kinetic energy ✓ Correct
B. The change in its potential energy
C. Its total mechanical energy
D. The work done by gravity alone
Solution: W_net = ΔKE = ½mv² − ½mu². It holds for ALL forces — conservative or not.
Q2 — Work Energy Theorem · medium · theory
If the net work done on a body during a displacement is zero, then over that displacement its:
A. Acceleration is zero throughout
B. Speed is unchanged ✓ Correct
C. Velocity is unchanged
D. Displacement is zero
Solution: W_net = ΔKE = 0 ⇒ same speed (direction may still change — e.g. uniform circular motion).
Q3 — Work Energy Theorem · easy · numerical
A 2 kg body accelerates from 5 m/s to 10 m/s. The net work done on it is:
A. 50 J
B. 75 J ✓ Correct
C. 25 J
D. 100 J
Solution: W = ½m(v² − u²) = 1×(100 − 25) = 75 J.
Q4 — Work Energy Theorem · medium · numerical
A constant net force of 10 N acts on a 2 kg body, initially at rest, over 4 m. Its final speed is:
A. 20 m/s
B. √40 ≈ 6.3 m/s ✓ Correct
C. 10 m/s
D. 4.5 m/s
Solution: W = Fs = 40 J = ½mv² ⇒ v = √(2×40/2) = √40 ≈ 6.3 m/s.
Q5 — Work Energy Theorem · hard · numerical
A car's braking force is constant. If its speed is doubled, its stopping distance becomes:
A. 4 times ✓ Correct
B. 2 times
C. √2 times
D. 8 times
Solution: Fd = ½mv² ⇒ d ∝ v². Doubling v quadruples the stopping distance — the key road-safety result.
Q6 — Work Energy Theorem · medium · numerical
A 1 kg body moving at 20 m/s is brought to rest by a constant opposing force of 50 N. The distance covered before stopping is:
A. 10 m
B. 8 m
C. 2 m
D. 4 m ✓ Correct
Solution: d = KE/F = (½×1×400)/50 = 200/50 = 4 m.
Q7 — Work Energy Theorem · medium · numerical
A 20 g bullet moving at 500 m/s penetrates 10 cm into a wooden block before stopping. The average resistive force is:
A. 25000 N ✓ Correct
B. 5000 N
C. 2500 N
D. 50000 N
Solution: KE = ½×0.02×250000 = 2500 J; F = KE/d = 2500/0.1 = 25000 N.
Q8 — Work Energy Theorem · easy · numerical
Starting from rest, a net work of 64 J is done on a 2 kg body. Its final speed is:
A. 32 m/s
B. 8 m/s ✓ Correct
C. 4 m/s
D. 16 m/s
Solution: ½mv² = 64 ⇒ v = √(64) = 8 m/s.