A 10 m long rope has linear density 0.5 kg/m. It lies length-wise on a smooth horizontal floor when a force of 25 N pulls it. Determine the tension in the rope at a point 6 m away from the point of application of force.

10 N

If total mass of the rope is m.

Since, the density is uniform, the left part has mass 0.6 m and the right part has mass 0.4 m.

As the rope moves as a single piece. So, both the parts have same acceleration.

Thus :

$T - T' = 0.6ma$

$T' = 0.4ma$

$T - T' = 0.6\frac{T'}{0.4}$

$T - T' = 1.5 T'$

$T = 2.5 T'$

$T' = \frac{2}{5}T$

Substitute T = 25 N; 

$T' = \frac{2}{5}\times 25$

  $= 10 N$