A uniform rope of mass M and length L is pulled by a constant force of 10 N. Find the tension (in newton) in the rope at a point at a distance L/5 from the end where the force is applied

Mass per unit length of the rope is $m=\frac{M}{L}$.  Let us find the tension at point P at a distance x from the end x=0.  Let T be the tension in the rope at point P.

For part AP the tension is towards the positive x-direction and for part BP the tension is towards the negative x-direction.  If a is the acceleration produced in the rope by the constant force F, then for part AP,

  T = (mass of AP) x a or T = mxa       (1)

 For part BP, we have

F-T = (mass of BP) x a = m(L-x)a    (2)

From (1), we have $a=\frac{T}{mx}$

 Using this in (2), we get

 $F-T=m(L-x)\times \frac{T}{mx}=\frac{\left(L-x\right)T}{x}$

Or     $F=T\left[\frac{\left(L-x\right)}{x}+1\right]=T\frac{L}{x}orT=\frac{Fx}{L}$

 At       $x=L-\frac{L}{5}=\frac{4L}{5}$,

   $T=\frac{F}{L}\times \frac{4L}{5}=\frac{4F}{5}=\frac{4\times 10}{5}=8N$