A homogeneous rope of mass per unit length λand length l kept on ground and one end of the rope

step 1 The momentum of the rope at any time is given by p is called to lambda y minus x into v. So forc

A homogeneous rope of mass per unit length λand length l...

A homogeneous rope of mass per unit length $\lambda$ and length $l$ kept on ground and one end of the rope is fixed to ground at $O .$ The left end of the rope (with respect to fixed end) is pulled by an external agent which imparts constant velocity $v$ to it. Find the work done by the external agent (in joule) to place the moving end extremely right with respect to fixed end. Take $\lambda=1 \mathrm{~kg} / \mathrm{m}, v=1 \mathrm{~ms}^{-1}$ and $l=1 \mathrm{~m}$

Key Concepts

Variable Mass Systems

This concept addresses situations in which the mass of a system is not constant over time. In problems where parts of a rope, chain, or any continuous medium are rapidly accelerated from rest as they join a moving segment, the effective mass in motion increases with time. Understanding variable mass systems is crucial, as it requires accounting for the momentum added by each new mass element when applying Newton’s laws and energy methods.

Work?Energy Principle

The work–energy principle states that the net work done on a system is equal to its change in kinetic energy. In the context of continuously pulling a rope with one end at rest, each infinitesimal segment must be accelerated from zero to the constant speed. The external agent’s work is thus accumulated as the sum (integral) of the kinetic energy imparted to each segment, ultimately resulting in a net kinetic energy proportional to the total mass and the square of the constant velocity.

Integration over a Continuous Mass Distribution

When dealing with a rope, which is a continuous distribution of mass, the energy or work contributions of infinitesimally small segments must be summed using integral calculus. This integration accounts for the fact that as the rope is pulled, new segments are sequentially accelerated. The total work done is calculated by integrating the small amounts of work required to accelerate each differential mass element from rest to the final velocity.

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