Physics · Work, Energy And Power · NEET
Because a net force causes acceleration (F = ma), and acceleration changes speed. When you push a body over a distance, you feed energy into its motion. Start from W = Fs = (ma)s. Now use v^2 = u^2 + 2as, which gives as = (v^2 - u^2)/2. Substituting: W = m(v^2 - u^2)/2 = (1/2)mv^2 - (1/2)mu^2 = K_f - K_i. So the work you do shows up exactly as a change in kinetic energy - nothing is lost in the derivation.
It is the NET work - the work done by the total (resultant) force. This is the most common mistake. If gravity, friction, and an applied push all act, you must add all their works: W_net = W_gravity + W_friction + W_applied. Only this sum equals K_f - K_i. The work of a single force alone does not equal the total kinetic energy change unless it is the only force acting.
No. The simple derivation using v^2 = u^2 + 2as assumes a constant force, but the theorem is fully general. For a variable force you derive it using integration: W = integral of F dx = integral of m(dv/dt)dx = integral of mv dv, which gives (1/2)mv^2 - (1/2)mu^2 again. NCERT proves this separately as 'the work-energy theorem for a variable force.' So the result W_net = K_f - K_i holds for any force.
Yes. If the net work is negative (force opposes motion, like friction or air resistance), the body slows down and K_f is less than K_i, so K_f - K_i is negative. Example: a braking car does negative work through friction, losing kinetic energy. Negative work means energy is taken out of the motion; positive work means energy is put in.
No, they are different. The work-energy theorem (W_net = delta K) is derived directly from Newton's second law and is always true, even with friction. Conservation of mechanical energy (KE + PE = constant) only holds when only conservative forces act. The work-energy theorem is more general - it counts the work of every force, including non-conservative ones like friction.
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 constant with a value 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 8x10^-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.
The work done by the net force acting on a body equals the change in its kinetic energy: W_net = K_f - K_i, where K = (1/2)mv^2.
W_net = (1/2)mv^2 - (1/2)mu^2, where u is the initial speed and v is the final speed of the body of mass m.
Start with W = Fs = (ma)s. Use v^2 = u^2 + 2as to get as = (v^2 - u^2)/2. Substitute: W = (1/2)mv^2 - (1/2)mu^2 = K_f - K_i.
Yes. Using integration, W = integral of F dx = integral of mv dv = (1/2)mv^2 - (1/2)mu^2, so the theorem holds for any force, constant or variable.
It lets you find final speed, stopping distance, or unknown forces without dealing with time or acceleration directly. Many NEET numericals (like the raindrop and circular-motion PYQs) are solved fastest with W_net = delta K.
