Work Done By A Constant Force

Work done by a constant force equals the product of the force magnitude, the displacement, and the cosine of the angle between them, written $W=Fd\cos \theta$. Only the component of force along the displacement transfers energy to the body; the perpendicular component does no work. Here the horizontal displacement is along the floor, so the relevant component of the 50 N force is $50\cos 60^{\circ }=25\text{ N}$. Multiplying by the 8 m displacement gives $W=25\times 8=200\text{ J}$. The option 400 J wrongly uses the full force without the cosine factor. The option 346 J mistakenly uses $\cos 30^{\circ }$, effectively swapping the angle. The option 100 J halves the correct displacement contribution and has no physical basis. This follows the NCERT definition of the scalar product of force and displacement. As a plausibility check, since the force is tilted away from the motion, the work must be less than the $400\text{ J}$ a fully horizontal force would deliver, and $200\text{ J}$ satisfies that bound.