Work-energy Theorem

The work-energy theorem asserts that the net work done on a body equals the change in its kinetic energy, $W_{net}=\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$. Since the trolley begins at rest, the initial kinetic energy is zero and the full 150 J converts into final kinetic energy. Setting $150=\frac{1}{2}(3)v_f^2$ gives $v_f^2=300/3=100$, so $v_f=10\text{ m/s}$. The option 100 m/s mistakes $v_f^2$ for $v_f$, skipping the square root. The option 5 m/s halves the correct speed by misapplying the one-half factor twice. The option 50 m/s arises from dividing work by mass without the kinetic-energy structure. It is worth emphasising that the theorem holds no matter how the force varies along the path, because only the total net work enters the balance, which makes it a powerful shortcut whenever the detailed force-versus-time history is unknown or messy. This is precisely how NCERT introduces the work-energy theorem for a constant net force. As a plausibility check, $\frac{1}{2}(3)(10{)}^2=150\text{ J}$ reproduces the input work exactly, confirming both magnitude and consistency.