Work by Constant Force — IISER Physics MCQs with Solutions
Free IISER Physics Work by Constant 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 Constant Force · easy · theory
The work done by a force is negative when the angle θ between the force and the displacement satisfies:
A. θ = 90° only
B. 0° ≤ θ < 90°
C. θ = 0° only
D. 90° < θ ≤ 180° ✓ Correct
Solution: W = Fs cosθ; cosθ < 0 for obtuse angles, so work is negative when the force has a component opposite to the displacement (e.g. friction).
Q2 — Work by Constant Force · medium · theory
A particle moves in a circle at constant speed. The work done by the centripetal force over any part of the path is:
A. Positive over half the circle, negative over the other half
B. Negative
C. Zero, because the force is always perpendicular to the velocity ✓ Correct
D. Positive
Solution: W = ∫F·ds; the centripetal force is always ⊥ to the instantaneous displacement, so cos90° = 0 ⇒ W = 0 everywhere on the path.
Q3 — Work by Constant Force · easy · numerical
A force of 20 N acts on a block at 60° to the horizontal while it moves 5 m horizontally. The work done by the force is:
A. 50 J ✓ Correct
B. 86.6 J
C. 100 J
D. 25 J
Solution: W = Fs cosθ = 20 × 5 × cos60° = 100 × 0.5 = 50 J. Trap: use the angle with the displacement, not the vertical.
Q4 — Work by Constant Force · medium · numerical
A force F⃗ = (3î + 4ĵ) N displaces a particle by d⃗ = (2î + 3ĵ) m. The work done is:
A. 12 J
B. 6 J
C. 17 J
D. 18 J ✓ Correct
Solution: W = F⃗·d⃗ = 3×2 + 4×3 = 6 + 12 = 18 J (dot product, not magnitudes multiplied).
Q5 — Work by Constant Force · hard · numerical
A 2 kg block is dragged 10 m along a floor (μ = 0.5) by a horizontal force of 15 N. The NET work done on the block is:
A. 50 J ✓ Correct
B. −100 J
C. 250 J
D. 150 J
Solution: Friction = μmg = 0.5×2×10 = 10 N. W_net = (15 − 10) × 10 = 50 J. (W_applied = 150 J, W_friction = −100 J.) Trap: net work uses the net force.
Q6 — Work by Constant Force · medium · numerical
A 5 kg body is lifted vertically through 4 m at constant velocity. The work done by gravity during the lift is:
A. −200 J ✓ Correct
B. Zero
C. −50 J
D. +200 J
Solution: W_gravity = −mgh = −5×10×4 = −200 J (gravity opposes the upward displacement). The lifting force does +200 J; net work = 0 (constant velocity).
Q7 — Work by Constant Force · hard · theory
A man holds a 20 kg load stationary on his shoulder for 5 minutes and gets tired. The work he does on the load, and the reason, are:
A. Zero — no displacement of the load; his fatigue is internal (muscular) energy expenditure ✓ Correct
B. Negative, since gravity acts down
C. ⁓10⁴ J, from the muscular effort
D. Positive, proportional to the holding time
Solution: Mechanical work needs displacement: W = Fs cosθ = 0 for s = 0. Physiological energy is consumed in repeated muscle-fibre contractions, but no work is done ON the load.
Q8 — Work by Constant Force · hard · numerical
Two forces F⃗₁ = (4î + ĵ) N and F⃗₂ = (−2î + 3ĵ) N act simultaneously on a particle displaced by d⃗ = (2î + 2ĵ) m. The total work done on the particle is:
A. 12 J ✓ Correct
B. 16 J
C. 8 J
D. 20 J
Solution: F⃗_net = (2î + 4ĵ); W = 2×2 + 4×2 = 12 J. (Equivalently W₁ + W₂ = 10 + 2 = 12 J — work adds like scalars.)