Physics · Work, Energy And Power · NEET
For a constant force, work W = F x, which is exactly the area of a rectangle (height F, width x). When force varies, NCERT splits the path into thin strips of width delta-x where force is nearly constant, so each strip's work is F times delta-x = area of that thin strip. Adding all strips gives the total area under the curve, which equals the total work. That is why 'area under F-x graph = work' is always true, whether the force is constant or variable.
For WORK, you always use the force-displacement (F-x) graph, because work = force times displacement. The area under a force-time (F-t) graph gives IMPULSE (change in momentum), not work. NEET loves this trap: read the x-axis label carefully. If the axis is position/displacement (metres), the area is work in joules. If the axis is time (seconds), the area is impulse in newton-second.
When the F-x curve goes below the x-axis, the force points opposite to the displacement, so that force does NEGATIVE work. You calculate that area as usual but subtract it. Total work = (area above axis) minus (area below axis). This is common when a force first pushes a body forward, then a resisting region slows it down.
Break the shape into rectangles, triangles and trapeziums. Rectangle area = base times height; triangle area = (1/2) times base times height; trapezium area = (1/2) times (sum of parallel sides) times width. Add the signed areas of every piece. Units: base in metres times height in newtons gives joules (J), the SI unit of work.
Yes, this is the standard NEET combo. The area under the F-x graph is the net work W. By the work-energy theorem, W = change in kinetic energy = (1/2)m v_final^2 minus (1/2)m v_initial^2. If the body starts from rest, W = (1/2)m v^2, so v = sqrt(2W/m). This is exactly how the NEET 2019 graph question is solved.
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.
Joule (J). Since area = newtons (force axis) times metres (displacement axis), and 1 newton times 1 metre = 1 joule, the graph area comes out directly in joules.
No. It works for both constant and variable forces. For a constant force the graph is a horizontal line and the area is a simple rectangle, giving W = F x. The graph method is just the general version.
They give the same answer. Integration W = integral of F dx from x_i to x_f is the mathematical way to find the area under the curve. The graph method is the geometric shortcut used when the graph is made of straight lines (rectangles, triangles, trapeziums).
NEET regularly asks graph-reading questions where you must find work from an F-x plot and then use the work-energy theorem to get speed. It tests two skills at once, so it is a high-value, frequently repeated pattern.
