Work Done By A Variable Force

For a force that varies with position, the work done equals the area under the force-versus-position graph, which generalises the constant-force result through integration $W=\int Fdx$. The graph here splits naturally into two pieces, so the total work is the sum of their areas. The first segment from $x=0$ to $x=2\text{ m}$ is a rectangle of area $10\times 2=20\text{ J}$. The second segment from $x=2$ to $x=4\text{ m}$ is a triangle of area $\frac{1}{2}(2)(10)=10\text{ J}$. Their sum is $20+10=30\text{ J}$. The option 40 J wrongly treats the falling part as a full rectangle. The option 20 J counts only the constant segment. The option 60 J doubles the total through an area error. This area method works for any force profile because work is fundamentally the accumulation of force over infinitesimal displacements, and summing those thin slivers is exactly what computing an area represents. The sign of each region also matters physically: a force opposing the motion would contribute negative area and reduce the total work delivered to the body. This is the NCERT graphical interpretation of work for a variable force. As a check, the triangular contribution must be smaller than the rectangular one, and $10<20$ confirms the breakdown.