Work by Tension in Circular Motion – Rankers Physics

Work Done by Constant and Variable Forces: Practice Problem & Solution

Assertion (A): A particle is rotated in a vertical circle with the help of a string. Work done by tension in the string on particle is zero. Reason (R): Tension is always perpendicular to instantaneous velocity.
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution Explained:

To solve this problem, we apply the core principles of Work Done by Constant and Variable Forces. Understanding the underlying formula is key to arriving at the correct answer below:

In circular motion, tension acts as the centripetal force, directed towards the center, while velocity is tangential. Thus, tension is always perpendicular to velocity (\(\theta = 90^\circ\)), meaning work done (\(W = Fd cos 90^\circ\)) is zero. Both are true, and Reason explains Assertion.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Karnam Malleshwari famous Indian weight lifter lifts a weight up and returns it to same initial position along the same path. Net work done by muscles of weight lifter is positive. Reason (R): Net displacement of weight is zero.  
2. Assertion (A): The work done during a round trip is always zero. Reason (R): No force is required to move a body in its round trip.
3. Two masses \(1\text{ g}\) and \(9\text{ g}\) are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is
4. Assertion (A): Power delivered by all forces acting on a particle moving in a uniform circular motion is always zero. Reason (R): Work done by all forces acting on a particle moving in a uniform circular motion is zero as KE remains constant.  
5. The position of a particle at any time \( t \) is given by \( x = t^2 + 1 \), where \( x \) is in m and \( t \) is in s. If a constant force of 8 N is acting on the particle, then the instantaneous power of the force at \( t = 1\text{ s} \) will be