Work By A Position-dependent Force

A force that depends on position requires integration to find its work, $W={\int }_0^{2}Fdx={\int }_0^{2}3x^{2}dx$. Evaluating the integral gives $W=[x^3{]}_0^2=2^3-0=8\text{ J}$. Since this is the only force, the work-energy theorem equates it to the gained kinetic energy: $8=\frac{1}{2}(1)v^2$, so $v^2=16$ and $v=4\text{ m/s}$. The option 8 m/s mistakes the work value in joules for the speed. The option 16 m/s reports $v^2$ rather than $v$. The option 2 m/s arises from forgetting to integrate and using $F$ at a single point. The need for integration arises because the force is not constant over the displacement; multiplying its value at any single point by the distance would systematically misrepresent the true accumulated work. Geometrically, the integral of $3x^2$ corresponds to the steadily growing area beneath a parabolic force curve, which weights later positions far more heavily than the early ones near the origin. This demonstrates the NCERT method of combining integration of a variable force with the work-energy theorem. As a check, the units of $\int 3x^{2}dx$ are $\text{N}\cdot \text{m}$, that is joules, and $\frac{1}{2}(1)(4{)}^2=8\text{ J}$ recovers the computed work.