A carpet of unit mass an unit length is laid on the floor. One end of the carpet is bent back and then pulled backwards with constants. Unit velocity, just above the part of the carpet which is still at rest on the floor. Find the speed of the centre of mass of the moving part. What is the minimum force needed to pull the moving part ?

Explanation

If the end of the folded part of the carpet is at a distance of `x` from the origin, the centre of mass of the folded part (which is moving) is at `3x//4` and hence has a speed of `0.75m//s`.
Now, force `=(dp)/(dt)=(d(mv))/(dt)=m((dv)/(dt))=0+v((dm)/(dt))=0.5N`
Note that the moving part of the carpet starts from the origin and the whole carpet will be moving when it reaches `x=2` after two units of time giving `((dm)/(dt))=0.5`
Note that the moving part of the carpet sets the next part into motion some energy is dissipated as heat.