Work by Constant Force — NEET Physics MCQs with Solutions
Free NEET 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. 90° < θ ≤ 180° ✓ Correct
B. θ = 0° only
C. 0° ≤ θ < 90°
D. θ = 90° only
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. Zero, because the force is always perpendicular to the velocity ✓ Correct
B. Negative
C. Positive
D. Positive over half the circle, negative over the other half
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. 100 J
B. 50 J ✓ Correct
C. 25 J
D. 86.6 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. 17 J
B. 6 J
C. 18 J ✓ Correct
D. 12 J
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. 150 J
B. 50 J ✓ Correct
C. 250 J
D. −100 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. −50 J
B. Zero
C. +200 J
D. −200 J ✓ Correct
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 · easy · numerical
A porter drags a 20 kg trunk 4 m across a platform with a 50 N force at 37° above the horizontal (cos37° = 0.8). The work done by the porter is:
A. 640 J
B. 200 J
C. 120 J
D. 160 J ✓ Correct
Solution: W = Fs cosθ = 50 × 4 × 0.8 = 160 J.
Q8 — Work by Constant Force · medium · numerical
A 10 kg suitcase is lifted from the floor onto a shelf 1.5 m high, then carried 10 m horizontally at the same height. The total work done against gravity is:
A. 150 J ✓ Correct
B. 1000 J
C. 1150 J
D. Zero
Solution: Against gravity: W = mgh = 10×10×1.5 = 150 J. The horizontal carry does no work against gravity (force ⊥ displacement) — the classic coolie trap.