The correct option is (c).
Explanation
Work is defined as the product of the component of force in the direction of the displacement and the magnitude of this displacement. Mathematically, for a constant force, Work ($W$) is the dot product of the Force vector ($\vec{F}$) and Displacement vector ($\vec{s}$):
$W = \vec{F} \cdot \vec{s} = Fs \cos \theta$
where $\theta$ is the angle between the force and the displacement.
Statement 1 is Correct:
The simplified formula $W = F \times s$ is valid specifically when the force is constant in magnitude and direction. If the force varies with position, the work done must be calculated using integration ($W = \int \vec{F} \cdot d\vec{s}$). Furthermore, this algebraic form implies that the force and displacement are in the same direction ($\cos 0^\circ = 1$).
Statement 2 is Incorrect:
When the force acts perpendicular to the direction of displacement, the angle $\theta$ is $90^\circ$. Since $\cos 90^\circ = 0$, the work done is zero ($W = 0$). The formula $W = F \times s$ would yield a non-zero value (assuming non-zero force and displacement), which contradicts the physical principle. Thus, the formula does not apply in this form for perpendicular forces.
Statement 3 is Correct:
When the displacement is in the opposite direction to the force (such as frictional force or retarding force), the angle $\theta$ is $180^\circ$. Since $\cos 180^\circ = -1$, the work done becomes:
$W = Fs \cos 180^\circ = -F \times s$
This correctly represents negative work.
Key Takeaway:
The expression $W = F \times s$ is a special case of the general work formula $W = Fs \cos \theta$, applicable only when a constant force acts parallel to the displacement. Work is zero if the force is perpendicular and negative if the force is opposite to the displacement.