Question

If \( \vec{F} \) and \( \vec{S} \) represent force and displacement respectively, then which of the following statements is true:

• A. Work is maximum when \( \vec{F} \) and \( \vec{S} \) are at right angles
• B. Work is positive when \( \vec{F} \) and \( \vec{S} \) are at obtuse angle
• C. Work is negative when \( \vec{F} \) and \( \vec{S} \) are at acute angle
• D. Work is zero if \( \vec{F} \) and \( \vec{S} \) are in opposite direction
• E. Area under \( \vec{F} \)-\( \vec{S} \) graph gives work done

Hint:

Work done is the product of force and displacement in the direction of the force. It can also be calculated as the area under the force-displacement graph.

Answer 1

Answer: (E)

The work done \( W \) by a force \( \vec{F} \) when it displaces an object through a displacement \( \vec{S} \) is given by: \[ W = \vec{F} \cdot \vec{S} = F S \cos \theta \] where:
- \( F \) is the magnitude of the force,
- \( S \) is the magnitude of the displaceent,
- \( \theta \) is the angle between the force and the displacement vectors.
Step 1: Work and angle between force and displacement.
- When \( \theta = 90^\circ \), \( \cos 90^\circ = 0 \), so work done is zero.
- When \( \theta = 0^\circ \), \( \cos 0^\circ = 1 \), so work is maximum.
- When \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \), so work is negative.

Step 2: Area under \( \vec{F} \)-\( \vec{S} \) graph.
The work done by a force is also represented by the area under the graph of force \( \vec{F} \) versus displacement \( \vec{S} \). This is the area under the curve on the graph, which gives the work done by the force.
Step 3: Conclusion.
Therefore, the correct statement is that the area under the \( \vec{F} \)-\( \vec{S} \) graph gives the work done, which corresponds to option (E).
Final Answer:} (E) Area under \( \vec{F} \)-\( \vec{S} \) graph gives work done