Question

A uniform thick rope of length $5\,m$ is kept on a frictionless surface and a force of $5\,N$ is applied to one of its ends. Find tension in the rope at $1\,m$ from this end.
A. $1\,N$
B. $3\,N$
C. $4\,N$
D. $5\,N$

Answer 1

Hint: In order to solve this question, we will first calculate mass per unit length of the rope and the net acceleration produced on the rope due to applied force and then we will use the concept that at any point on the rope the tension force will be responsible for motion of remaining part of the rope to calculate tension force.

Formula used:
Force applied on a body is $F$ and $m$ is the mass of a body then acceleration $a$ is given by Newton’s second law of motion.
$a = \dfrac{F}{m}$

Complete step by step answer:
According to the question, we have given that
Total length of rope, $l = 5m$
Applied force on the rope, $F = 5N$
Let ‘m’ be the total mass of the rope then mass per unit length m’ will be
$m' = \dfrac{m}{l} \\
\Rightarrow m'= \dfrac{m}{5}$
let a be the acceleration produced in the rope then using,
$a = \dfrac{F}{m}$
$\Rightarrow a = \dfrac{5}{m}$

Now, we need to calculate tension at a distance of $1m$ from the end of applied force, then this tension force T will produce acceleration of remaining part of rope which will be of length $l - 1 = 5 - 1 = 4m$
So, tension force T will move the net mass of $4m$ length which can be calculated as
$M = m'l'$ where M is the mass of $4m$ length and $l' = 4$ put $m' = \dfrac{m}{5}$ we get,
$\Rightarrow M = \dfrac{{4m}}{5}$

Now tension force $T$ will produce same acceleration of this part of rope as same as of complete rope which we have already calculated as $a = \dfrac{5}{m}$ so we have,
$T = Ma$
On putting values we get,
$T = \dfrac{{4m}}{5} \times \dfrac{5}{m}$
$ \therefore T = 4N$

Hence, the correct option is C.

Note: It should be remembered that, mass of any part of the rope is simply the product of length of that part and mass per unit length of the rope and here tension force at any point on the rope will be of different value but the sum of tension forces due to two parts of rope will always be equal to net force applied on the rope.