Physics · Work, Energy And Power · NEET
Yes. This is the key point for NEET. Whether the force is constant or variable, the theorem W = K_f - K_i is always true. For a constant force you compute W = F d cos theta; for a variable force you compute W = integral of F dx (the area under the force-displacement graph). The right side, change in kinetic energy, stays exactly the same. NCERT proves it starting from dK/dt = m v (dv/dt) = F v = F (dx/dt), which gives dK = F dx, and integrating both sides gives K_f - K_i = integral of F dx = W.
Three steps. Step 1: find the work done W = area under the F-x graph between the start and end positions (add positive areas, subtract areas below the x-axis where force is negative). Step 2: this area equals the change in kinetic energy, so W = (1/2) m v_f^2 - (1/2) m v_i^2. Step 3: solve for v_f. If the body starts from rest, v_i = 0, so v_f = square root of (2W/m). You never need the exact force formula, only the area.
They are the same operation. If the problem gives you a formula like F = 20 + 10y, integrate: W = integral of F dy. If the problem gives you a graph of F versus x, find the geometric area (triangles and rectangles). Both give the work done, and both equal the change in kinetic energy. Use whichever the question hands you.
NCERT says the work-energy theorem is an integral form of Newton's second law. Newton's second law F = m a tells you the force at one instant. When you multiply by dx and add up (integrate) over the whole path, the acceleration information turns into a change in kinetic energy. So the theorem does not give the full instant-by-instant motion, but it directly connects total work to the change in speed, which is often all you need.
No. The work is still in joules and kinetic energy is still (1/2) m v^2 in joules. Only the method of getting W changes (integration or area instead of a simple multiplication). The final relation W = change in KE is identical.
A force F = 20 + 10y acts on a particle in the y-direction, where F is in newton and y is in metre. The work done by this force to move the particle from y = 0 to y = 1 m is:
An object of mass 500 g, initially at rest, is acted upon by a variable force whose X-component varies with X as shown in the graph. The velocities of the object at the points X = 8 m and X = 12 m would be, respectively (nearly):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The net work done by a variable force on a body equals the change in its kinetic energy: W = integral of F dx = K_f - K_i, where K = (1/2) m v^2. It is valid for any force, constant or variable.
Start with the rate of change of kinetic energy: dK/dt = d/dt[(1/2)m v^2] = m v (dv/dt). By Newton's second law m (dv/dt) = F, so dK/dt = F v = F (dx/dt). This gives dK = F dx. Integrating from initial to final position: K_f - K_i = integral of F dx = W.
It is the integral form of Newton's second law. Newton's law relates force and acceleration at each instant; the work-energy theorem sums the effect over a path, so it does not carry the full instant-by-instant information but directly links total work to change in speed.
Yes. If the net work (net signed area under the F-x graph, or the integral of F dx) is negative, the kinetic energy decreases and the body slows down. This happens whenever the force acts opposite to the displacement over part of the path.
NEET regularly gives problems where you must find work by integrating a position-dependent force or by taking the area under an F-x graph, then use W = change in KE to get the final speed. It rewards students who remember that the theorem holds even when the force is not constant.
