Work by Constant Force — JEE Main Physics MCQs with Solutions
Free JEE Main 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. 90° < θ ≤ 180° ✓ Correct
C. θ = 0° only
D. 0° ≤ θ < 90°
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. Positive
C. Positive over half the circle, negative over the other half
D. Negative
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. 86.6 J
C. 25 J
D. 50 J ✓ Correct
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. 18 J ✓ Correct
B. 6 J
C. 12 J
D. 17 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. 50 J ✓ Correct
B. 150 J
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. +200 J
B. −50 J
C. −200 J ✓ Correct
D. Zero
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 · medium · numerical
A force F⃗ = (2î − ĵ + k̂) N moves a particle from (1, 2, 3) m to (4, 4, 2) m. The work done is:
A. 9 J
B. 5 J
C. 7 J
D. 3 J ✓ Correct
Solution: d⃗ = (3, 2, −1); W = 2×3 + (−1)×2 + 1×(−1) = 6 − 2 − 1 = 3 J.
Q8 — Work by Constant Force · hard · numerical
A 2 kg block is pushed 5 m up a 37° incline (μ = 0.25, sin37° = 0.6, cos37° = 0.8) at constant speed by a force along the incline. The work done by the applied force is:
A. 60 J
B. 100 J
C. 40 J
D. 80 J ✓ Correct
Solution: F = mg sinθ + μmg cosθ = 12 + 4 = 16 N. W = 16 × 5 = 80 J (60 J against gravity + 20 J against friction).