Work by Variable Force — JEE Main Physics MCQs with Solutions
Free JEE Main 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 slope of the F–x graph
B. Always zero
C. F × (x₂ − x₁) using any F
D. The area under the F–x graph between x₁ and x₂ ✓ Correct
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
B. −½kx² (spring force opposes the stretch) ✓ Correct
C. Zero
D. −kx²
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. 18 J
B. 6 J
C. 9 J ✓ Correct
D. 12 J
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. 25 J
B. 70 J
C. 45 J ✓ Correct
D. 20 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. 35 J
B. 30 J
C. 50 J
D. 40 J ✓ Correct
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. 0.5 J
B. 1.0 J
C. 1.5 J ✓ Correct
D. 2.0 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 · medium · numerical
A force F = 3x² (N) moves a particle from x = 1 m to x = 2 m. The work done is:
A. 8 J
B. 9 J
C. 7 J ✓ Correct
D. 12 J
Solution: W = ∫₁² 3x² dx = [x³]₁² = 8 − 1 = 7 J.
Q8 — Work by Variable Force · hard · numerical
A force F = A/x² acts on a particle moving along x. The work done in moving from x = a to x = 2a is:
A. A/2a ✓ Correct
B. A ln 2
C. A/a
D. 3A/2a
Solution: W = ∫ₐ²ᵃ A x⁻² dx = A[−1/x]ₐ²ᵃ = A(1/a − 1/2a) = A/2a. Trap: the ln form belongs to F ∝ 1/x, not 1/x².