Question

If \(\vec{F}\) and \(\vec{S}\) represent the applied force and displacement of an object, then the work done is

• A. zero if \(\vec{F}\) and \(\vec{S}\) are in the same direction
• B. maximum if \(\vec{F}\) and \(\vec{S}\) are at right angles to each other
• C. the area under the graph between \(\vec{F}\) and \(\vec{S}\)
• D. positive if the angle between \(\vec{F}\) and \(\vec{S}\) is obtuse
• E. negative if the angle between \(\vec{F}\) and \(\vec{S}\) is acute

Hint:

To remember the sign of work:
Same direction (\(0^\circ\)) \(\to\) Positive.
Opposite direction (\(180^\circ\)) \(\to\) Negative.
Perpendicular (\(90^\circ\)) \(\to\) Zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Work done (\(W\)) by a force is defined as the dot product of the force vector (\(\vec{F}\)) and the displacement vector (\(\vec{S}\)).
It can be expressed as \(W = F S \cos \theta\), where \(\theta\) is the angle between the force and displacement.

Step 3: Detailed Explanation:
- Option A: If \(\vec{F}\) and \(\vec{S}\) are in the same direction, \(\theta = 0^\circ\). Since \(\cos 0^\circ = 1\), work is maximum, not zero.
- Option B: If they are at right angles, \(\theta = 90^\circ\). Since \(\cos 90^\circ = 0\), work is zero, not maximum.
- Option C: For a variable force, work is calculated by the integral \(\int F \cdot dS\). Geometrically, this represents the area under the force-displacement graph.
- Option D: If the angle is obtuse (\(90^\circ < \theta \le 180^\circ\)), \(\cos \theta\) is negative, so work is negative.
- Option E: If the angle is acute (\(0^\circ \le \theta < 90^\circ\)), \(\cos \theta\) is positive, so work is positive.

Step 4: Final Answer:
The work done is represented by the area under the force-displacement graph.