Question

A particle moving in a horizontal circle of radius 0.5 m completes half rotation. The work done by the centripetal force of 5 N on the particle (in J) is:

• A. 2
• B. 5
• C. 2.5
• D. 3
• E. 0

Hint:

Centripetal force is a "no-work" force because it is always perpendicular to the direction of motion.

Answer 1

Answer: (E)

Step 1: Concept
Work done ($W$) is defined by the dot product of force ($\vec{F}$) and displacement ($\vec{d}$), given by $W = Fd \cos \theta$, where $\theta$ is the angle between the force and displacement vectors.

Step 2: Direction of Force
In circular motion, the centripetal force always acts toward the center of the circle (along the radius).

Step 3: Direction of Displacement
The instantaneous displacement of the particle is always tangential to the circular path.

Step 4: Conclusion
Since the radius is always perpendicular to the tangent, the angle $\theta$ is $90^\circ$. Because $\cos 90^\circ = 0$, the work done by the centripetal force is zero, regardless of the distance traveled. Final Answer: (E)