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.
Answer & explanation
Correct answer: option 2
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;
$T' = \frac{2}{5}\times 25$
$= 10 N$