Work by Variable Force — NEET Physics MCQs with Solutions
Free NEET Physics Work by Variable Force 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 by Variable Force · easy · theory
For a force that varies with position, the work done between x₁ and x₂ equals:
A. The area under the F–x graph between x₁ and x₂ ✓ Correct
B. Always zero
C. F × (x₂ − x₁) using any F
D. The slope of the F–x graph
Solution: W = ∫F dx = area under the force–displacement curve (areas below the axis count negative).
Q2 — Work by Variable Force · medium · theory
A spring is stretched by x from its natural length. The work done BY the spring force during this stretch is:
A. −½kx² (spring force opposes the stretch) ✓ Correct
B. +½kx²
C. −kx²
D. Zero
Solution: The spring force (−kx) is opposite to the displacement while stretching, so W_spring = −½kx²; the external agent does +½kx².
Q3 — Work by Variable Force · easy · numerical
A force F = 2x (N) acts on a particle. The work done in moving it from x = 0 to x = 3 m is:
A. 12 J
B. 18 J
C. 6 J
D. 9 J ✓ Correct
Solution: W = ∫₀³ 2x dx = [x²]₀³ = 9 J. Trap: don't use F(3)×3 = 18 J — the force varies.
Q4 — Work by Variable Force · medium · numerical
A force F = (4 + 2x) N acts along x. The work done from x = 0 to x = 5 m is:
A. 45 J ✓ Correct
B. 20 J
C. 70 J
D. 25 J
Solution: W = ∫₀⁵ (4 + 2x)dx = 4×5 + [x²]₀⁵ = 20 + 25 = 45 J.
Q5 — Work by Variable Force · hard · numerical
An F–x graph rises linearly from 0 to 10 N over x = 0 → 2 m, then stays constant at 10 N up to x = 5 m. The total work done is:
A. 50 J
B. 40 J ✓ Correct
C. 30 J
D. 35 J
Solution: Triangle: ½×2×10 = 10 J; rectangle: 10×3 = 30 J. Total = 40 J (area under the curve).
Q6 — Work by Variable Force · medium · numerical
A spring of constant k = 100 N/m is already stretched 0.1 m. The additional work needed to stretch it from 0.1 m to 0.2 m is:
A. 1.5 J ✓ Correct
B. 2.0 J
C. 1.0 J
D. 0.5 J
Solution: W = ½k(x₂² − x₁²) = 50(0.04 − 0.01) = 1.5 J. Trap: NOT ½k(x₂ − x₁)² = 0.5 J.
Q7 — Work by Variable Force · easy · numerical
The work required to stretch a spring of k = 400 N/m by 5 cm from its natural length is:
A. 0.5 J ✓ Correct
B. 0.05 J
C. 1.0 J
D. 10 J
Solution: W = ½kx² = ½ × 400 × (0.05)² = 200 × 0.0025 = 0.5 J.
Q8 — Work by Variable Force · medium · numerical
A spring stores 4 J when stretched by 2 cm. The energy stored when it is stretched by 4 cm is:
A. 4 J
B. 32 J
C. 8 J
D. 16 J ✓ Correct
Solution: U = ½kx² ∝ x². Doubling x quadruples the energy: 4 × 4 = 16 J.