NDA (II) 2016 General Ability Test · Question 104 of 150

Correct — B, it increases. The work-energy theorem says that the net work done on a body equals the change in its kinetic energy, written as W = ΔK = K_final − K_initial. Read that equation backwards and the answer is immediate: if W is a positive number, then K_final − K_initial is positive, so the final kinetic energy is larger than the initial one. Physically, positive net work means the resultant force has a component pointing the same way the body is moving, so it pushes the body along and speeds it up.

• (a)Decreases — This is what happens when the net work is negative — a resultant force opposing the motion, such as friction on a sliding block or the brakes on a car. Negative work removes kinetic energy; the question specifies positive work.
• (c)Remains constant — Kinetic energy stays put only when the net work is zero, which happens either when there is no resultant force or when the resultant force is always perpendicular to the motion. A satellite in a circular orbit and a car cruising at a steady speed are the standard examples.
• (d)Becomes zero — Kinetic energy falls to zero only when the body is brought to rest, which needs enough negative work to cancel all the energy the body started with. Positive work moves it in the opposite direction entirely.

Work is done when a force acting on a body produces a displacement, and it is measured by the component of the force along the displacement multiplied by that displacement. It is a scalar and its SI unit is the joule. The work-energy theorem links it to motion: whatever the forces are and however complicated the path, the algebraic sum of the work done by all of them equals the change in the body's kinetic energy, K = ½mv².

The theorem is the fastest route to a large family of NDA problems, because it lets you jump from forces straight to speeds without touching acceleration or time. The sign convention is the whole of this item — positive work when force and displacement point the same way, negative when they oppose, zero when they are perpendicular. That last case explains two results students often find surprising: the tension in the string of a body whirled in a circle does no work at all, and neither does the normal reaction on a block sliding along a horizontal floor.

• The work-energy theorem states that the net work done on a body equals the change in its kinetic energy.
• Work is the product of the displacement and the component of the force along it, so a force perpendicular to the motion does no work.
• The SI unit of both work and energy is the joule, equal to one newton-metre.
• Kinetic energy is ½mv², is always positive or zero, and is a scalar.

• Forgetting that the theorem needs the NET work, that is the work of all forces added algebraically, not the work of one convenient force.
• Assuming that a force always does work — a perpendicular force does none, however large it is.

The GAT asks this either as a one-line sign question, as here, or as a short numerical where you equate the work done against friction to the kinetic energy lost.

• practice — not a real PYQ

  A body is moved along a horizontal floor. The work done on it by the normal reaction from the floor is

  • (a)positive
  • (b)negative
  • (c)zero
  • (d)equal to the work done by friction

  Answer(c) zero — the normal reaction is perpendicular to the displacement, and a perpendicular force does no work.

• practice — not a real PYQ

  A car of mass 1000 kg moving at 20 m/s is brought to rest by the brakes. The work done by the braking force is

  • (a)+200 kJ
  • (b)−200 kJ
  • (c)+400 kJ
  • (d)−400 kJ

  Answer(b) −200 kJ — the kinetic energy falls from ½ × 1000 × 20² = 200 kJ to zero, so the net work done is a loss of 200 kJ.