A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is
- (a)zero
- (b)positive
- (c)infinite
- (d)negative
Correct — D, negative. Work done W = F × d × cos θ, where θ is the angle between the force and the displacement. Here the force is anti-parallel to (exactly opposite) the displacement, so θ = 180° and cos 180° = −1, making the work negative. A force acting against the motion removes energy from the body, so the work it does is negative.
- (a)zero — Work is zero only when the force is perpendicular to the displacement (θ = 90°, cos 90° = 0), not when it is anti-parallel.
- (b)positive — Work is positive when the force has a component along the displacement (θ < 90°); an anti-parallel force acts against the motion, so its work is negative.
- (c)infinite — For a finite force acting over a finite displacement the work is a finite number, never infinite.
Work is a scalar quantity given by W = F·d·cos θ, the product of the force, the displacement, and the cosine of the angle between them. Its sign is set entirely by that angle: positive for θ below 90°, zero at 90°, and negative for θ above 90° (up to 180°, directly opposing the motion).
'Anti-parallel to its displacement' is the key phrase — it fixes θ at 180°, so cos θ = −1 and the work must be negative. Friction and other opposing forces do negative work in the same way.
- W = F·d·cos θ; the sign depends only on the angle θ between force and displacement.
- θ = 0° gives maximum positive work; θ = 90° gives zero work; θ = 180° gives negative work.
- A force opposing the motion (like friction) does negative work and removes kinetic energy.
- Work is measured in joules (J) and is a scalar, so it has sign but no direction.
The force here is anti-parallel to the displacement, so θ = 180° and the work done is negative.
- Assuming any applied force does positive work — it depends on the direction relative to the motion.
- Confusing the anti-parallel case (θ = 180°, negative) with the perpendicular case (θ = 90°, zero).
Asked as the sign of work when the force and the displacement are anti-parallel (opposite) — cos 180° gives a negative answer.
No directly related past PYQ was found.
- practice — not a real PYQ
The work done by the force of friction on a block sliding across a floor is
- (a)positive
- (b)zero
- (c)negative
- (d)infinite
Answer(c) negative — friction acts opposite to the motion, so it does negative work.
- practice — not a real PYQ
A body moves in a circle at constant speed. The work done by the centripetal force is
- (a)positive
- (b)negative
- (c)zero
- (d)maximum
Answer(c) zero — the centripetal force is perpendicular to the velocity, so cos 90° = 0.