When a stone tied to a string is whirled in a circle, the work done on it by the string:

Correct — C, is zero. Work is not simply force times distance; it is the component of the force along the displacement, W = Fd cos θ. The string can only pull, and it pulls along its own length towards the hand at the centre. The stone, meanwhile, moves along the circle, and the tangent to a circle is perpendicular to its radius. So the force and the displacement are at right angles at every instant, cos 90° is zero, and the string does no work over any part of the path, not merely over a complete revolution. The energy accounting says the same thing from the other side. By the work-energy theorem the net work done on a body equals the change in its kinetic energy, and the string does not change the stone's speed — it only bends the velocity round, turning the direction without touching the magnitude. That is what a centripetal force does for a living. The point holds even for a stone whirled in a vertical circle, where the speed plainly does change between the bottom and the top: that change is gravity's work, not the string's, and the tension in an inextensible string stays perpendicular to the motion throughout.

• (a)is positive. — Positive work would speed the stone up, since it would add kinetic energy. The string does nothing of the kind — it pulls sideways to the motion, never along it.
• (b)is negative. — Negative work would slow the stone down, which is what a force acting against the displacement does. The tension acts across the displacement, not against it.
• (d)depends on the mass of the stone. — Mass changes how much tension is needed to hold the stone on its circle, but not the angle between that tension and the motion. The angle stays 90° whatever the stone weighs, so the work stays zero.

A body moving in a circle at steady speed is accelerating, because its velocity is changing direction even though its magnitude is fixed. The acceleration points at the centre and has magnitude v²/r, and the force that supplies it — tension in a string, gravity for a satellite, friction for a car on a bend, the electrostatic pull on an orbiting electron in the old atomic model — is called the centripetal force. Because it is always at right angles to the velocity, a centripetal force never does work and never changes kinetic energy.

The item tests whether a candidate has internalised the cosine in the definition of work rather than the phrase 'force times distance'. Three angles are worth committing to memory: force along the motion gives positive work, force against it gives negative work, force across it gives none. The third is the one papers keep coming back to, because the situations are so common — a satellite circling the Earth, a porter walking on level ground with a load on his head, the moon in its orbit. Note also what the question is careful to ask. It asks about the work done BY THE STRING, not the total work done on the stone. In a horizontal circle those are the same thing; in a vertical circle they are not, and the difference is gravity.

• Work is W = Fd cos θ, so a force perpendicular to the displacement does no work.
• Tension in the string acts along the radius; the stone's displacement is along the tangent; the two are always at right angles.
• By the work-energy theorem, zero net work means no change in kinetic energy, which matches the constant speed of uniform circular motion.
• The centripetal acceleration is v²/r and is directed at the centre; it changes the direction of the velocity, not its magnitude.
• A satellite in a circular orbit is the same case — gravity supplies the centripetal force and does no work on it.

• Reading work as force times distance and forgetting the cosine.
• Assuming the work is zero only over a full revolution; it is zero at every instant, because the angle is always a right angle.
• Confusing the work done by the string with the total work done on the stone when the circle is vertical.

As this qualitative item, as a numerical using W = Fd cos θ, or as a statements question about which forces in a listed situation do no work.

• practice — not a real PYQ

  A satellite moves in a circular orbit around the Earth. The work done on it by the Earth's gravitational force in one complete revolution is:

  • (a)positive and equal to its kinetic energy
  • (b)negative and equal to its potential energy
  • (c)zero
  • (d)dependent on the mass of the satellite

  Answer(c) zero — gravity supplies the centripetal force and acts at right angles to the satellite's motion throughout a circular orbit, so it does no work.

• practice — not a real PYQ

  A force of 10 N acts on a body that moves 5 m in a direction making an angle of 60° with the force. The work done is:

  • (a)50 J
  • (b)25 J
  • (c)43 J
  • (d)0 J

  Answer(b) 25 J — W = Fd cos θ = 10 × 5 × cos 60° = 10 × 5 × 0.5 = 25 J.