Physics · Work, Energy And Power · NEET
You must use the NET work, meaning the sum of work done by every force. The theorem is W(net) = Kf - Ki. If a box is pushed with 50 N while friction resists with 20 N over 4 m, W(net) = (50 - 20)(4) = 120 J, not 50(4). A common mistake is using only the applied force and forgetting friction or gravity. When one force alone acts, then W(net) equals that force's work.
Compute the net work W, then write W = (1/2)m v^2 - (1/2)m u^2 and solve for v. Rearranged, v = sqrt(u^2 + 2W/m). Example: a 2 kg body at rest (u = 0) gains 100 J of net work, so 100 = (1/2)(2)v^2, giving v^2 = 100 and v = 10 m/s. This works even when the force is not constant, as long as you know the total work done.
Friction acts opposite to the motion, so the angle between force and displacement is 180 degrees and cos 180 = -1. This makes W(friction) = -F d, a negative value. In a stopping problem the car's kinetic energy falls to zero, so W(net) = 0 - Ki, which is negative. Setting -F d = -Ki gives F d = Ki, the neat result that the braking work removes all the starting kinetic energy.
Use the work-energy theorem when the force is variable, when several forces act, or when the path is curved (like a circle or an incline). Kinematics equations only work for constant acceleration along a straight line. The theorem still gives the right speed even if acceleration changes, because it deals with total energy, not step-by-step motion. For simple constant-force straight-line cases both give the same answer.
Yes. Kinetic energy depends only on speed, not direction, so the theorem holds on any path. For a particle speeding up on a circle, the net work equals the change in (1/2)m v^2. In the NEET 2016 circular-motion PYQ, the tangential force does work over the arc length travelled (two revolutions), and that work equals the gained kinetic energy. The centripetal force does zero work because it is always perpendicular to the velocity.
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If F(A) and F(B) are the braking forces on cars A and B respectively, then the ratio F(A)/F(B) is
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g = 10 m/s^2. The work done by the (i) gravitational force and the (ii) resistive force of air is
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. The magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10^-4 J by the end of the second revolution after the beginning of the motion, is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W(net) = Kf - Ki = (1/2)m v^2 - (1/2)m u^2, where W(net) is the total work by all forces, u is initial speed and v is final speed.
Yes. If the body slows down, Kf is less than Ki, so W(net) is negative. This happens when resistive forces like friction or air drag do more negative work than any positive work.
Yes. The theorem holds for constant and variable forces. For a variable force you find the total work by integration or as the area under a force-displacement graph, then equate it to the change in kinetic energy.
Many NEET problems ask for final speed, stopping distance, or an unknown force. The theorem skips finding acceleration and time, giving a fast one-step link between force, distance and speed - a big time saver in the exam.
Positive. A falling body moves in the same direction as gravity, so W(grav) = m g h is positive. For a body rising, gravity does negative work equal to -m g h.
