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a particle moves in such a way that the line AB translates parellel to itself.the displacement dr a of particke A is equal to the displacement dr b of the particle B in any short time interval.the net work done by the internal force Fab and Fba is? 2)what is the sum of work done by Fab and Fba? 3)is there would be any change in question if AB does not translate parellel to itself but rotates?

Profile image of Krish Mishra
15 Years agoGrade Upto college level
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Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To tackle your question about the work done by internal forces in a system of particles, let's break it down step by step. We have two particles, A and B, connected in such a way that they can exert forces on each other, denoted as Fab (force exerted by A on B) and Fba (force exerted by B on A). The scenario describes a system where the line AB translates parallel to itself, and we need to analyze the work done by these internal forces.

Understanding Work Done by Internal Forces

In physics, the work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this can be expressed as:

  • Work = Force × Displacement × cos(θ)

Here, θ is the angle between the force and the displacement vector. In our case, since the line AB translates parallel to itself, the forces Fab and Fba act along the line connecting A and B.

1. Net Work Done by Internal Forces

Since the displacements of particles A and B are equal in magnitude and direction (dr_a = dr_b), we can analyze the work done by each force:

  • Work done by Fab = Fab × dr_a
  • Work done by Fba = Fba × dr_b

However, according to Newton's third law, the forces exerted by A on B and by B on A are equal in magnitude and opposite in direction:

  • Fab = -Fba

When we calculate the net work done by both forces, we find:

  • Net Work = Work done by Fab + Work done by Fba
  • Net Work = (Fab × dr_a) + (Fba × dr_b)
  • Net Work = (Fab × dr_a) + (-Fab × dr_a) = 0

Thus, the net work done by the internal forces Fab and Fba is zero.

2. Sum of Work Done by Fab and Fba

As we just calculated, the sum of the work done by Fab and Fba is:

  • Sum of Work = Work done by Fab + Work done by Fba = 0

This reinforces the idea that internal forces do not contribute to the net work done on the system as a whole, which is a fundamental principle in mechanics.

3. Impact of Rotation on the System

If the line AB does not translate parallel to itself but instead rotates, the situation changes significantly. In this case, the forces exerted by A and B may not act along the same line of action, which means that the angle θ between the force and the displacement could vary. This would affect the work done by each force:

  • Work done by Fab = Fab × dr_a × cos(θ_a)
  • Work done by Fba = Fba × dr_b × cos(θ_b)

In a rotating system, the internal forces can do work on the particles, leading to a change in kinetic energy or potential energy depending on the configuration. Therefore, the net work done by the internal forces could be non-zero, potentially resulting in changes in the system's energy state.

In summary, when AB translates parallel to itself, the net work done by internal forces is zero. However, if AB rotates, the situation becomes more complex, and the internal forces can contribute to the work done, affecting the overall energy of the system.